Chapter 14.3 - Contrapositive And Converse

In the previous section, we saw quantifiers. In this section, we will see Implications.

Implications

• In mathematics, we often encounter statements with:
    ♦ “if-then”
    ♦ “only if”
    ♦ “if and only if”
Let us see some examples:
Example 1
This can be written in 4 steps:
1. Consider the following statement:
r: If you are born in some country, then you are a citizen of that country.
2. We can see that, r is a compound statement. It’s component statements are:
p: You are born in some country.
q: You are citizen of that country.
3. Now r becomes: if p then q
• “if p then q” has a definite meaning.
We can write:
    ♦ Whenever p is true,
    ♦ q is also true.
• We can also write:
    ♦ Whenever p is false,
    ♦ We must discard the compound statement r.
4. An important point to remember is:
“if p then q” does not give any guarantee that p happens.

Example 2
This can be written in 4 steps:
1. Consider the following statement:
r: If a number is a multiple of 9, then it is a multiple of 3.
2. We can see that, r is a compound statement. It’s component statements are:
p: A number is a multiple of 9.
q: That number is a multiple of 3.
3. Now r becomes: if p then q
• “if p then q” has a definite meaning.
We can write:
    ♦ Whenever p is true,
    ♦ q is also true.
• For our present example:
    ♦ Whenever a number is a multiple of 9
    ♦ That number will be a multiple of 3 also.
• We can also write:
    ♦ Whenever p is false,
    ♦ We do not know whether q is true or false.
• For our present example:
    ♦ Whenever a number is not a multiple of 9
    ♦ We must discard the compound statement r.
4. An important point to remember is:
“if p then q” does not give any guarantee that p happens.
• For our present example:
Just by reading r, we cannot say that a number is a multiple of 9.
• r is a link between p and q.
• p cannot stand alone independently.

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Based on example 2, we can write some more details. It can be written in 5 steps:
1. When we are given “if p then q”, we can write it in another form:
"p implies q"
• Symbolically, we denote this as: p ⇒ q
    ♦ The symbol "⇒" stands for “implies”.
• For our present example, we can write:
    ♦ A number is a multiple of 9.
    ♦ implies that
    ♦ That number is a multiple of 3.
2. When we are given “if p then q”, we can write:
"p is a sufficient condition for q".
• For our present example, we can write:
    ♦ Knowing a number to be a multiple of 9,
    ♦ is sufficient
    ♦ To know that, the number is a multiple of 3.
3. When we are given “if p then q”, we can write:
"p only if q".
• For our present example, we can write:
    ♦ A number is a multiple of 9
    ♦ only if
    ♦ That number is a multiple of 3.
4. When we are given “if p then q”, we can write:
"q is a necessary condition for p"
• For our present example, we can write:
    ♦ For a number to be a multiple of 9
    ♦ it is necessary that
    ♦ That number is a multiple of 3   
5. When we are given “if p then q”, we can write:
"~q implies ~p"
• For our present example, we can write:
    ♦ When a number is not a multiple of 3
    ♦ it implies that
    ♦ That number is not a multiple of 9.   
◼ When we are given “if p then q”, we can rewrite it in the above five different ways.

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Contrapositive statement

• When we are given “if p then q”, we can write:
“if ~q then ~p”.
• This is known as the contrapositive of “if p then q”.
• Note that, the order of p and q is changed in the contrapositive statement.
• Let us see some examples:

Example 1:
This can be written in 4 steps:
1. Consider the statement:
r: If a number is divisible by 9, then it is divisible by 3.
2. The component statements are:
p: A number is divisible by 9.
q: That number is divisible by 3.
3. Let us write the negation statements:
~p: A number is not divisible by 9.
~q: That number is not divisible by 3.
4. Now we can write “if ~q then ~p”:
If a number is not divisible by 3 then that number is not divisible by 9.
• This is the contrapositive of r.

Example 2:
This can be written in 4 steps:
1. Consider the statement:
r: If you are born in India, then you are a citizen of India.
2. The component statements are:
p: You are born in India.
q: You are a citizen of India.
3. Let us write the negation statements:
~p: You are not born in India.
~q: You are not a citizen of India.
4. Now we can write “if ~q then ~p”:
If you are not a citizen of India, then you are not born in India.
• This is the contrapositive of r.

Example 3:
This can be written in 4 steps:
1. Consider the statement:
r: If a triangle is equilateral, it is isosceles.
2. The component statements are:
p: A triangle is equilateral.
q: That triangle is isosceles.
3. Let us write the negation statements:
~p: A triangle is not equilateral.
~q: That triangle is not isosceles.
4. Now we can write “if ~q then ~p”:
If a triangle is not isosceles, then that triangle is not equilateral.
• This is the contrapositive of r.

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Converse statement

• When we are given “if p then q”, we can write:
“if q then p”.
• This is known as the converse of “if p then q”.
• Note that, the order of p and q is changed in the converse statement.
• Let us see some examples:

Example 1:
This can be written in 3 steps:
1. Consider the statement:
r: If a number is divisible by 10, then it is divisible by 5.
2. The component statements are:
p: A number is divisible by 10.
q: That number is divisible by 5.
3. Now we can write “if q then p”:
If a number is divisible by 5, then that number is divisible by 10.
• This is the converse of r.

Example 2:
This can be written in 3 steps:
1. Consider the statement:
r: If a number n is even, then n² is even.
2. The component statements are:
p: A number n is even.
q: The square of that number n² is even.
3. Now we can write “if q then p”:
If the number n² is even, then n is even.
• This is the converse of r.

Example 3:
This can be written in 3 steps:
1. Consider the statement:
r: If you do all the exercises in the book, then you get an A grade in the class.
2. The component statements are:
p: You do all the exercises in the book.
q: You get an A grade in the class.
3. Now we can write “if q then p”:
If you get an A grade in the class, then you have done all the exercises in the book.
• This is the converse of r.

Example 4:
This can be written in 3 steps:
1. Consider the statement:
r: If two integers a and b are such that a > b, then (a-b) is always a positive integer.
2. The component statements are:
p: Two integers a and b are such that a > b.
q: (a-b) is always a positive integer.
3. Now we can write “if q then p”:
If two integers a and b are such that (a-b) is always a positive integer, then a > b.
• This is the converse of r.

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Let us see a solved example:

Solved example 14.8
For each of the following compound statements, first identify the corresponding component statements. Then check whether the statements are true or not.
(i) If a triangle ABC is equilateral, then it is isosceles.
(ii) If a and b are integers, then ab is a rational number.
Solution:
Part (i):
1. Consider the statement:
r: If a triangle ABC is equilateral, then it is isosceles.
2. The component statements are:
p: A triangle ABC is equilateral.
q: That triangle is isosceles.
3. An equilateral triangle has all three sides equal. So obviously, the two sides other than the base will be equal. That means, an equilateral triangle is an isosceles triangle also. 
• We can write:
If p happens, q is true. So r is true.

Part (ii):
1. Consider the statement:
r: If a and b are integers, then ab is a rational number.
2. The component statements are:
p: a and b are integers.
q: ab is a rational number.
3. Integers do not have any decimal portions. So the product of two integers will not have any decimal portions. That means, the product will be an integer rational number. An integer is a rational number.
• We can write:
If p happens, q is true. So r is true.

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“If an only if” statement 

This can be explained in steps:
1. In some cases, we encounter the condition: “p if and only if q”
2. Let us see an example:
• A student can apply for MSc Maths if he has taken Maths, Physics and Computer science for the Bsc course.
• A student can apply for MSc Chemistry if and only if he has taken Maths, Physics and Chemistry for the Bsc course.
3. In some other cases, both the conditions below need to be satisfied:
    ♦ “p if and only if q”
    ♦ “q if and only if p”
• In such cases we use the symbol "⇔"
    ♦ Using this symbol, we can write: p ⇔ q.
4. Note that:
(i) “p if and only if q”
    ♦ is different from
    ♦ p ⇔ q.
(ii) “q if and only if p”
    ♦ is different from
    ♦ p ⇔ q.
(iii) “p if and only if q”
    ♦ is different from
    ♦ “q if and only if p”
5. If we are given "p ⇔ q", then all the four conditions below must be satisfied:
(i) p if and only if q
(ii) q if and only if p
(iii) p is necessary and sufficient condition for q
(iv) q is necessary and sufficient condition for p

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Let us see a solved example:
Solved example 14.9
Given below are pairs of statements. Combine them using “if and only if”.
(i) p: If a rectangle is a square, then all it’s four sides are equal.
q: If all the four sides of a rectangle are equal, then that rectangle is a square.

(ii) p: If the sum of digits of a number is divisible by 3, then the number is divisible by 3.
q: If a number is divisible by 3, then the sum of it’s digits is divisible by 3.

(iii) p: If a tumbler is half empty, then it is half full.
q: If a tumbler is half full, then it is half empty.

Solution:
Part (i)
1. Consider the statement:
p: If a rectangle is a square, then all it’s four sides are equal.
• This p can be considered as a "if r then s" statement.
• So the component statements are:
    ♦ r: A rectangle is a square.
    ♦ s: All four sides of that rectangle are equal.
• It is clear that "if r then s" is applicable.
2. Consider the statement:
q: If all the four sides of a rectangle are equal, then that rectangle is a square.
• This q contains r and s that we saw above. It can be considered as a "if s then r" statement.
• So the component statements are:
    ♦ s: All the four sides of a rectangle are equal.
    ♦ r: That rectangle is a square.
• It is clear that "if s then r" is applicable.
3. So both the two conditions below are applicable:
    ♦ "if r then s"
    ♦ "if s then r"
4. Since both the conditions are applicable, we can combine r and s using "if and only if". We get:
A rectangle is a square if and only if all four sides of that rectangle are equal.

Part (ii)
1. Consider the statement:
p: If the sum of digits of a number is divisible by 3, then the number is divisible by 3.
• This p can be considered as a "if r then s" statement.
• So the component statements are:
    ♦ r: The sum of digits of a number is divisible by 3.
    ♦ s: The number is divisible by 3.
• It is clear that "if r then s" is applicable.
2. Consider the statement:
q: If a number is divisible by 3, then the sum of it’s digits is divisible by 3.
• This q contains r and s that we saw above. It can be considered as a "if s then r" statement.
• So the component statements are:
    ♦ s: A number is divisible by 3.
    ♦ r: The sum of it’s digits is divisible by 3.
• It is clear that "if s then r" is applicable.
3. So both the two conditions below are applicable:
    ♦ "if r then s"
    ♦ "if s then r"
4. Since both the conditions are applicable, we can combine r and s using "if and only if". We get:
The sum of digits of a number is divisible by 3 if and only if the number is divisible by 3.

Part (iii)
1. Consider the statement:
p: If a tumbler is half empty, then it is half full.
• This p can be considered as a "if r then s" statement.
• So the component statements are:
    ♦ r:  A tumbler is half empty.
    ♦ s: That tumbler is half full.
• It is clear that "if r then s" is applicable.
2. Consider the statement:
q: If a tumbler is half full, then it is half empty.
• This q contains r and s that we saw above. It can be considered as a "if s then r" statement.
• So the component statements are:
    ♦ s: A tumbler is half full.
    ♦ r: That tumbler is half empty.
• It is clear that "if s then r" is applicable.
3. So both the two conditions below are applicable:
    ♦ "if r then s"
    ♦ "if s then r"
4. Since both the conditions are applicable, we can combine r and s using "if and only if". We get:
A tumbler is half empty if and only if that tumbler is half full.

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The link to some solved examples is given below:

Exercise 14.4

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In the next section, we will see validating statements.

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Chapter 14.2 - Special Words And Quantifiers

In the previous section, we saw compound statements and component statements. In this section, we will see Special words/phrases.

Special words/phrases

• We saw that, component statements are connected by words like “and”, “or” etc.,
• These words are known as connectives.
• When used in mathematical statements, these words have special meanings. We need to clearly understand those special meanings.

The word “And”

This can be explained in 3 steps:
1. First we will see an example. It can be written in 4 steps:
(i) Given below, is a compound statement with the word “and”:
p: A point occupies a position and it’s location can be determined.
(ii) The component statements are:
q: A point occupies a position.
r: It’s location can be determined.
   ♦ q and r are connected by “and”.
(iii) Checking component statements:
• q is true.
• r is true.
(iv) Since both q and r are true, p is true.
2. Let us see another example. It can be written in 4 steps:
(i) Given below, is a compound statement with the word “and”:
p: 42 is divisible by 5, 6 and 7.
(ii) The component statements are:
q: 42 is divisible by 5.
r:  42 is divisible by 6.
s:  42 is divisible by 7.
   ♦ q, r and s are connected by “and”.
(iii) Checking component statements:
• q is false.
• r is true.
• s is true
(iv) Since q is false, p is false.
(Even though r and s are true, p is false because one component q is false)
3. We can write two rules related to “and”:
(i) A compound statement with “and” is true if all it’s component statements are true.
(ii) A compound statement with “and” is false if one or more of it’s component statements is false.

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Let us see a solved example
Solved example 14.5
Write the component statements of the following compound statements and check whether the compound statements are true or false.
(i) A line is straight and extends indefinitely in both directions.
(ii) 0 is less than every positive integer and every negative integer.
(iii) All living things have two legs and two eyes.
Solution:
Part (i):
1. We have:
p: A line is straight and extends indefinitely in both directions.
2. The component statements are:
q: A line is straight.
r: A line extends indefinitely in both directions.
   ♦ q and r are connected by “and”.
3. Checking component statements:
• q is true
• r is true.
4. Since both q and r are true, p is true.

Part (ii):
1. We have:
p: 0 is less than every positive integer and every negative integer.
2. The component statements are:
q: 0 is less than every positive integer.
r: 0 is less than every negative integer.
   ♦ q and r are connected by “and”.
3. Checking component statements:
• q is true
• r is false.
4. Since r is false, p is false.

Part (iii):
1. We have:
p: All living things have two legs and two eyes.
2. The component statements are:
q: All living things have two legs.
r: All living things have two eyes.
   ♦ q and r are connected by “and”.
3. Checking component statements:
• q is false
• r is false.
4. Since one or more of the component statements is false, p is false. In this case both the components q and r are false.

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Let us see an interesting case involving “and”. It can be written in 4 steps:
1. Consider the sentence:
A mixture of alcohol and water can be separated by chemical methods.
2. We are tempted to write two separate sentences:
(i) A mixture of alcohol can be separated by chemical methods.
(ii) A mixture of water can be separated by chemical methods.
3. Let us analyze the two sentences:
(i) The sentence in 2(i) does not have any meaning. This is because, if alcohol alone is present, we cannot call it a ‘mixture’. If there is no ‘mixture’, there is no need for separating.
(ii) The sentence in 2(ii) does not have any meaning. This is because, if water alone is present, we cannot call it a ‘mixture’. If there is no ‘mixture’, there is no need for separating.
4. It is clear that, the word “and” in this case is not used for connecting two statements. It is used for connecting two substances alcohol and water.

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The word “Or”

This can be explained in 6 steps:
1. First we will see an example. It can be written in 4 steps:
(i) Given below, is a compound statement with the word “or”:
p: Two lines in a plane, either intersect at a point or they are parallel.
(ii) The component statements are:
q: Two lines in a plane, intersect at a point.
r: Two lines in a plane, are parallel.
   ♦ q and r are connected by “or”.
(iii) Checking component statements:
• if q is true, then r is false.
• if r is true, then q is false.
(iv) One of the components will be always true.
If one of the components (q or r) is true, p is true.
2. Let us see another example. It can be written in 3 steps:
(i) Given below, is another compound statement with the word “or”:
p: 125 is a multiple of 7 or 8.
(ii) The component statements are:
q: 125 is a multiple of 7.
r: 125 is a multiple of 8.
   ♦ q and r are connected by “or”.
(iii) Checking component statements:
• q is false.
• r is false.
(iv) If all the components are false, p is false.
3. Let us see one more example. It can be written in 3 steps:
(i) Given below, is a compound statement with the word “or”:
p: The school is closed if there is a holiday or sunday.
(ii) The component statements are:
q: The school is closed if there is a holiday.
r: The school is closed if it is sunday.
   ♦ q and r are connected by “or”.
(iii) Checking component statements:
• q is true.
• r is true.
(iv) If one of the components is true, p is true.
4. We can write two rules related to “or”:
(i) A compound statement with “or" is true if one or more of it’s component statements are true.
(ii) A compound statement with “or” is false if all of it’s component statements are false.
5. The word "or" is used in two ways in English language. This can be explained using two examples.

Example 1:
This can be written in 3 steps:
(i) Consider the compound statement with word "or":
p: An ice cream or pepsi is available with lunch in a restaurant.
(ii) The component statements are:
q: An ice cream is available with lunch in a restaurant.
r: A pepsi is available with lunch in a restaurant.
   ♦ q and r are connected by “or”.
(iii) Let us analyze the component statements:
• If a customer does not want pepsi, he can opt for ice cream.
• If a customer does not want ice cream, he can opt for pepsi.
• Only one item (ice cream or pepsi) will be allowed.
• The “or” used in such cases is called exclusive “or”.

Example 2:
This can be written in 3 steps:
(i) Given below, is a compound statement with the word “or”:
p: A student who has taken biology or chemistry can apply for M.Sc. microbiology program.
(ii) The component statements are:
q: A student who has taken biology can apply for M.Sc. microbiology program.
r: A student who has taken chemistry can apply for M.Sc. microbiology program.
   ♦ q and r are connected by “or”.
(iii) Let us analyze the component statements:
• If a student has taken biology, he/she can apply for M.Sc. microbiology program.
• If a student has taken chemistry, he/she can apply for M.Sc. microbiology program.
• If a student has taken both biology and chemistry, he/she can apply for M.Sc. microbiology program.
• The “or” used in such cases is called inclusive “or”.
6. We must clearly understand the difference between exclusive “or” and inclusive “or”. Then only we will be able to check whether a compound statement is true or false.

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Now we will see some solved examples.

Solved example 14.6
For each of the following statements, determine whether an inclusive “or” or exclusive “or” is used. Give reasons for your answer.
(i) To enter a country, you need a passport or a voter registration card.
(ii) The school is closed if it is a holiday or a Sunday.
(iii) Two lines intersect at a point or are parallel.
(iv) Students can take French or Sanskrit as their third language.
Solution:
Part (i):
1. We have:
p: To enter a country, you need a passport or a voter registration card.
2. The component statements are:
q: To enter a country, you need a passport.
r: To enter a country, you need a voter registration card.
   ♦ q and r are connected by “or”.
3. Let us analyze the component statements:
• If a person has a passport, he/she can enter a country.
• If a person has a voter registration card, he/she can enter a country.
• If a person has both passport and voter registration card, he/she can enter a country.
• So the “or” used in this case is inclusive “or”.

Part (ii):
1. We have:
p: The school is closed if it is a holiday or a Sunday.
2. The component statements are:
q: The school is closed if it is a holiday.
r: The school is closed if it is a Sunday.
   ♦ q and r are connected by “or”.
3. Let us analyze the component statements:
• On a holiday, the school is closed.
• On a Sunday, the school is closed.
• On a day, if Sunday and holiday coincide, then also the school is closed.
• So the “or” used in this case is inclusive “or”.

Part (iii):
1. We have:
p: Two lines intersect at a point or are parallel.
2. The component statements are:
q: Two lines in a plane, intersect at a point.
r: Two lines in a plane, are parallel.
   ♦ q and r are connected by “or”.
3. Let us analyze the component statements:
• if q is true, then r is false.
• if r is true, then q is false.
4. Both cannot be true at the same time.
• Both cannot be false at the same time.
• So the “or” used in this case is exclusive “or”.

Part (iv):
1. We have:
p: Students can take French or Sanskrit as their third language.
2. The component statements are:
q: Students can take French as their third language.
r: Students can take Sanskrit as their third language.
   ♦ q and r are connected by “or”.
3. Let us analyze the component statements:
• if q is true, then r is false.
• r is true, then q is false.
• No student can take both languages at the same time.
• So the “or” used in this case is exclusive “or”.

Solved example 14.7
For each of the following statements, determine whether an inclusive “or” or exclusive “or” is used. Give reasons for your answer.
(i) √2 is a rational number or an irrational number.
(ii) To enter into a public library children need an identity card from the school or a letter from the school authorities.
(iii) A rectangle is a quadrilateral or a 5-sided polygon.
Solution:
Part (i):
1. We have:
p: √2 is a rational number or an irrational number.
2. The component statements are:
q: √2 is a rational number.
r: √2 is an irrational number.
   ♦ q and r are connected by “or”.
3. Let us analyze the component statements:
• if q is true, then r is false.
• if r is true, then q is false.
4. Both cannot be true at the same time.
• Both cannot be false at the same time.
• So the “or” used in this case is exclusive “or”.
5. In this case, r is true. So p is true.

Part (ii):
1. We have:
p: To enter into a public library children need an identity card from the school or a letter from the school authorities.
2. The component statements are:
q: To enter into a public library children need an identity card from the school.
r: To enter into a public library children need a letter from the school authorities.
   ♦ q and r are connected by “or”.
3. Let us analyze the component statements:
• If identity card is available, entry is possible.
• If letter is available, entry is possible.
• If both identity card and letter is available, entry is possible.
• So the “or” used in this case is inclusive “or”.

Part (iii):
1. We have:
p: A rectangle is a quadrilateral or a 5-sided polygon.
2. The component statements are:
q: A rectangle is a quadrilateral.
r: A rectangle is a 5-sided polygon.
   ♦ q and r are connected by “or”.
3. Let us analyze the component statements:
• if q is true, then r is false.
• if r is true, then q is false.
(Recall that, a quadrilateral has only four sides)
4. Both cannot be true at the same time.
• Both cannot be false at the same time.
• So the “or” used in this case is exclusive “or”.
5. In this case, q is true. So p is true.

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Quantifiers

Some basics can be written in 4 steps:
1. Consider the set of all rectangles.
• There will be infinite number of rectangles in that set.
2. We know that, squares are also rectangles.
• So some of the members in the set will be squares.
3. This fact can be written as a statement:
p: There exists a rectangle whose all sides are equal.
• This statement means that, there is at least one rectangle whose all sides are equal.
4. The phrase “There exist” is a quantifier.
• Quantifiers are phrases which give us an idea about the “quantity of items”
• “There exists” gives us the idea that, there is at least one item.

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Let us see another quantifier. It can be explained in steps:
1. Let S be the set of all prime numbers.
• Then we can write S in the set-builder form as:
S = {p : p is a prime number}
2. Now, square root of any prime number will be an irrational number.
• We want to write this as a statement. We can write it as:
q: For every prime number p, √p is an irrational number.
3. The phrase “For every” is a quantifier.
• As we mentioned just above, Quantifiers are phrases which give us an idea about the “quantity of items”
• “For every” gives us the idea that, all items in the set are to be taken into consideration.
• In the previous example, we had to consider only some members of the set  (members which are squares).

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It is important to consider the position of the quantifier. This can be explained using an example. It can be written in 5 steps:
1. Consider two sentences:
(i) For every positive number x there exists a positive number y such that y < x
(ii) There exists a positive number y such that for every positive number x, we have y < x
(In both the sentences, “number” means, real number)
2. Let us analyze the first sentence.
• We first consider the set P of positive numbers:
P = {x : x is a +ve number}
• The sentence says that:
We can pick any x from that set. We will be able to find a positive number y such that, y is less than x.
• For example, (y= x/2) will be less than x.
3. Let us analyze the second sentence.
• We first consider the set P of positive numbers:
P = {y : y is a +ve number}
• The sentence says that:
We can pick any y from that set. That y will be less than any positive number x.
• This is not possible. For example, y will be greater than (x= y/2).
• So this sentence is false.
4. Both sentences appear to be the same.
• But we see that:
   ♦ First sentence is true.
   ♦ Second sentence is false.
5. Why is that so?
• In the first sentence, the quantifier “there exists” is at the middle.
• In the second sentence, “there exists” is at the beginning.
• The quantifiers must be carefully introduced in a sentence. Each quantifier must be introduced precisely at the right place. Not too early and not too late.

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The link to some solved examples is given below:

Exercise 14.3

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In the next section, we will see implications.

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Chapter 14.1 - Negation of A Statement

In the previous section, we saw what a "statement" is. In this section, we will see new statements from old.

Some basic points can be written in 3 steps:
1. We are discussing about mathematical reasoning.
• We saw “statements” which are the basic units of mathematical reasoning.
• We want to know how statements help us in the process of mathematical reasoning.
2. The English mathematician George Boole put forward an effective technique to analyze statements.
• In this technique, we ask two questions:
   ♦ What happens when the statement is true ?
   ♦ What happens when the statement is false ?
3. Finding answers to the above two questions is a major step in mathematical reasoning.

---

So our next aim is to learn about Negation of a statement. This can be written in 3 steps:
1. The denial of a statement is called the negation of the statement.
2. Let us see an example. It can be written in 3 steps:
(i) Consider the following statement:
p: New Delhi is a city.
(ii) The negation of this statement can be written in any one of the three ways shown below:
• It is not the case that New Delhi is a city
• It is false that New Delhi is a city
• New Delhi is not a city.
(iii) If p is a statement, then negation of p is also a statement.
   ♦ It is denoted as ~p.
   ♦ It is read as ‘not p’.
3. ~p helps us to analyze p in a better way. This can be demonstrated using some examples.
Example 1:
This can be written in 4 steps:
(i) Consider the statement:
p: Every one in Germany speaks German.   
(ii) We want ~p.
• p tells us that all people living in Germany speak German.
• ~p must tell us that all people living in Germany does not speak German.
(iii) This ~p can be achieved by saying that there is at least one person in Germany, who does not speak that language.
• So we get:
~p: There is at least one person in Germany who does not speak German.
(iv) p and ~p enable us to think of various possibilities.

Example 2:
This can be written in 4 steps:
(i) Consider the statement:
p: Both the diagonals of a rectangle have the same length.   
(ii) We want ~p.
• p tells us that: Take any rectangle. Both it's diagonals will be of the same length.
• ~p must tell us that the statement is not true for any rectangle.
(iii) This ~p can be achieved by saying that there is at least one rectangle where both diagonals are not the same.
• So we get:
~p: There is at least one rectangle where both diagonals are not of the same length.
(iv) p and ~p enable us to think of various possibilities.

Example 3:
This can be written in 4 steps:
(i) Consider the statement:
p: √7 is rational.   
(ii) We want ~p.
• p tells us that: √7 is rational.
• ~p must tell us that the statement is not true.
(iii) This ~p can be achieved by saying that it is not rational.
• So we get:
~p: √7 is not rational.
(iv) p and ~p enable us to think of various possibilities.

---

Now we will see some solved examples.

Solved example 14.2
Write the negation of the following statements and check whether the resulting statements are true.
(i) Australia is a continent.
(ii) There does not exist a quadrilateral which has all it’s sides equal.
(iii) The sum of 3 and 4 is 9.
Solution:
Part (i):
1. Consider the statement:
p: Australia is a continent.
2. The negation can be written as:
~p: It is false that Australia is a continent.
OR
~p: Australia is not a continent.
3. ~p is false.

Part (ii):
1. Consider the statement:
p: There does not exist a quadrilateral which has all it’s sides equal.
2. We want ~p.
• p tells us that: Not even a single quadrilateral (with the given property) exists.
• ~p must tell us that "not even a single" is false.
3. This ~p can be achieved by saying that there is at least one quadrilateral.
• So we get:
~p: There is at least one quadrilateral which has all it’s sides equal.
4. ~p is true. (square is a quadrilateral which has all it’s sides equal)

Part (iii):
1. Consider the statement:
p: The sum of 3 and 4 is 9.
2. The negation can be written as:
~p: It is false that the sum of 3 and 4 is 9.
OR
~p: sum of 3 and 4 is not 9.
3. ~p is true.

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Compound statements

This can be explained in 4 steps:
1. Some mathematical statements are obtained by combining two or more simple statements.
• The simple statements are combined by using connecting words like “and”, “or” etc.,
2. This can be demonstrated using an example. It can be written in 2 steps:
(i) Consider the following statement:
p: There is something wrong with the bulb or with the wiring.
• This statement tells us that:
   ♦ The bulb is not turning on because,
   ♦ there is something wrong with the bulb.
   ♦ or there is something wrong with the wiring.
(ii) So the given statement is in fact formed by two statements:
q: There is something wrong with the bulb.
r: There is something wrong with the wiring.
   ♦ q and r are connected by “or”.
3. Another example:
(i) Consider the two statements:
p: 7 is an odd number.
q: 7 is a prime number.
• These two simple statements can be connected by the word “and”.
• We get:
r: 7 is both an odd number and a prime number.
4. A compound statement is a statement which is made up of two or more simple statements. Each of the simple statements is called a component statement.

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Let us see some solved examples:
Solved example 14.3
Find the component statements of the following compound statements.
(i) The sky is blue and the grass is green.
(ii) It is raining and it is cold.
(iii) All rational numbers are real and all real numbers are complex.
(iv) 0 is a positive number or a negative number.
Solution:
Part (i):
1. We have:
p: The sky is blue and the grass is green.
2. The component statements are:
q: The sky is blue.
r: The grass is green.
   ♦ q and r are connected by “and”.  

Part (ii):
1. We have:
p: It is raining and it is cold.
2. The component statements are:
q: It is raining.
r: It is cold.
   ♦ q and r are connected by “and”.  

Part (iii):
1. We have:
p: All rational numbers are real and all real numbers are complex.
2. The component statements are:
q: All rational numbers are real.
r:  All real numbers are complex.
   ♦ q and r are connected by “and”.

Part (iv):
1. We have:
p: 0 is a positive number or a negative number.
2. The component statements are:
q: 0 is a positive number.
r: 0 is a negative number.
   ♦ q and r are connected by “or”.

Solved example 14.4
Find the component statements of the following compound statements. Check each of the component statements and find whether they are true or false.
(i) A square is a quadrilateral and it’s four sides equal.
(ii) All prime numbers are either even or odd
(iii) A person who has taken Mathematics or Computer science can go for MCA
(iv) Chandigarh is the capital of Haryana and UP.
(v) 2 is a rational number or an irrational number.
(vi) 24 is a multiple of 2, 4 and 8.
Solution:
Part (i):
1. We have:
p: A square is a quadrilateral and it’s four sides equal.
2. The component statements are:
q: A square is a quadrilateral.
r: A square has all four sides equal.
   ♦ q and r are connected by “and”.
3. Checking component statements:
• q is true.
• r is also true.

Part (ii):
1. We have:
p: All prime numbers are either even or odd
2. The component statements are:
q: All prime numbers are even numbers.
r: All prime numbers are odd numbers.
   ♦ q and r are connected by “and”.
3. Checking component statements:
• q is false.
• r is false.

Part (iii):
1. We have:
p: A person who has taken Mathematics or Computer science can go for MCA.
2. The component statements are:
q: A person who has taken Mathematics can go for MCA.
r: A person who has taken Computer science can go for MCA.
   ♦ q and r are connected by “or”.
3. Checking component statements:
• q is true.
• r is true.

Part (iv):
1. We have:
p: Chandigarh is the capital of Haryana and UP.
2. The component statements are:
q: Chandigarh is the capital of Haryana.
r: Chandigarh is the capital of UP.
   ♦ q and r are connected by “and”.
3. Checking component statements:
• q is true.
• r is false.

Part (v):
1. We have:
p: √2 is a rational number or an irrational number.
2. The component statements are:
q: √2 is a rational number.
r: √2 is an irrational number.
   ♦ q and r are connected by “or”.
3. Checking component statements:
• q is false.
• r is true.

Part (vi):
1. We have:
p: 24 is a multiple of 2, 4 and 8.
2. The component statements are:
q: 24 is a multiple of 2.
r: 24 is a multiple of 4.
s: 24 is a multiple of 8.
   ♦ q, r and s are connected by “and”.
3. Checking component statements:
q is true.
• r is true.
• s is true

---

• We see that:
Component statements are connected by words like “and”, “or” etc., These words have special meanings in mathematics. We will see it in the next section.

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The link to a few more solved examples is given below:

Exercise 14.2

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In the next section, we will see special words/phrases.

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Chapter 14 - Mathematical Reasoning

In the previous section, we completed a discussion on limits and derivatives. In this chapter, we will see mathematical reasoning.

Some basics about reasoning can be written in 4 steps:
1. Human beings have the ability to engage in the process of reasoning.
• The "process of reasoning" is the process in which we think (in a logical way) about some thing. We think about some thing to arrive at a conclusion.
2. Let us see some examples:
• Early human beings thought about the “cause of lightning”. Other animals are not able to engage in such a thought.
• Early human beings thought about “why the pebbles found in river beds are rounded ?”. Other animals are not able to engage in such a thought.
• Human beings at present are thinking about “what is inside a proton ?”.  Other animals are not able to engage in such a thought.
3. The process of thinking must be done in a logical and sensible way. Then only we can call it reasoning.
4. On seeing a light above a distant hill, a person may think for a while and come to the conclusion that, the light is caused by some super natural power.
• Such thinking process cannot be called reasoning.
• We must investigate and analyze all available facts. Also, we must try to obtain new facts.

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• Reasoning can be done in mathematical problems also.
• We have see inductive reasoning in chapter 4. In this chapter we will see deductive reasoning.
• The basic unit required for mathematical reasoning is a statement.
 

• Some basics about “statement” in mathematics can be written in 6 steps
1. Consider the following three sentences:
(i) In 2021, the President of USA was a woman.
(ii) A blue whale weighs more than an elephant.
(iii) This building is beautiful.
2. Let us analyze each sentence:
(i) When we read the first sentence, we immediately decide that, the sentence is false.
No doubt arises in our mind. That is., we can write with out any doubt that, the sentence is false.     
(ii) When we read the second sentence, we immediately decide that, the sentence is true.
No doubt arises in our mind. That is., we can write with out any doubt that, the sentence is true.
(iii) When we read the third sentence, some doubt can arise.
    ♦ Some people may consider the building to be beautiful.
    ♦ Some other people may consider the building to be ordinary.
    ♦ Yet some others may consider the building to be ugly.
• For a person who consider the building to be beautiful, the sentence is true.
• For a person who consider the building to be ordinary or ugly, the sentence is false.
• In such cases,
    ♦ We cannot say that the sentence is true.
    ♦ We cannot say that the sentence is false.
• That means:
There is a possibility that, the sentence is both true and false.
• In such situations, we say that:
    ♦ the sentence is ambiguous.
    ♦ or we say that: there is ambiguity in the sentence.
(The words ambiguous or ambiguity refers to the situation where a sentence has more than one possible meaning)
3. Now we can write a basic idea about statements:
• If there is no ambiguity in a sentence, then that sentence can be considered as a statement in mathematics.
• If there is ambiguity in a sentence, then that sentence is not considered as a statement in mathematics.
4. Let us consider the three sentences that we wrote in step (1).
• Sentences (i) and (ii) can be considered as statements in mathematics.
• Sentence (iii) is not considered as a statement in mathematics.
5. Let us write a precise definition for statement. It can be written in 3 steps:
(i) A sentence is called a mathematically acceptable statement if it is true.
(ii) A sentence is called a mathematically acceptable statement if it is false.
(iii) A sentence is not called a mathematically acceptable statement if it is both true and false.
6. From now on wards, whenever we say “statement”, it should be understood that, it is a mathematically acceptable statement.

---

Let us see a few more examples:
1. Consider the sentence:
Three plus five equal eight.
• This sentence is true. There is no ambiguity. So it is a statement.
2. Consider the sentence:
The sum of two positive numbers is positive.
• This sentence is true. There is no ambiguity. So it is a statement.
3. Consider the sentence:
All prime numbers are odd numbers.
• This sentence is false. There is no ambiguity. So it is a statement.  
(Recall that there is one and only one even prime number, which is ‘2’)
4. Consider the sentence:
The sum of x and y is greater than zero.
• On some occasions, this sentence is true.
   ♦ For example, when x = 3 and y = 8
• On some other occasions, this sentence is false.
   ♦ For example, when x = 3 and y = -8
• There is ambiguity in this sentence. So it is not a statement.
5. Consider the sentence:
For any two natural numbers x and y, the sum (x+y) is greater than zero.  
• This sentence is true. There is no ambiguity. So it is a statement.  
(Recall that natural numbers are: 1, 2, 3, 4, . . .)
6. Consider the sentence:
How beautiful !
• This is an exclamatory sentence. Exclamatory sentences convey strong emotion or excitement. They are not considered as statements in mathematics.
7. Consider the sentence:
Open the door.
• This sentence is an order. Orders are not considered as statements in mathematics.
8. Consider the sentence:
Where are you going ?
• This sentence is a question. Questions are not considered as statements in mathematics.
9. Consider the sentence:
Tomorrow is Friday.
• On some occasions (Thursdays), this sentence is true.
• On some other occasions (days other than Thursdays), this sentence is false.
• There is ambiguity in this sentence. So it is not a statement.
◼ Word “tomorrow” indicates variable time. It can be any day of the week. It depends on the day on which the sentence is written. Similar is the case with the words “today”,  and “yesterday”. If any of these three words are present in a sentence, then it cannot be considered as a statement in mathematics.
10. Consider the sentence:
Kashmir is far from here.
• On some occasions, this sentence is true.
For example, when the sentence is written at a place far away from Kashmir.
• On some other occasions, this sentence is false.
For example, when the sentence is written at a place close to Kashmir.
• There is ambiguity in this sentence. So it is not a statement.
◼ Word “here” indicates variable position. It can be any position in space. It depends on the position at which the sentence is written. Similar is the case with the word “there”. If any of these two words are present in a sentence, then it cannot be considered as a statement in mathematics.
11. Consider the sentence:
She is a mathematics graduate.
• On some occasions, this sentence is true.
For example, when the person represented by the pronoun “she” is a mathematics graduate.
• On some other occasions, this sentence is false.
For example, when the person represented by the pronoun “she” is not a mathematics graduate .
• There is ambiguity in this sentence. So it is not a statement.
◼ Pronoun “she” indicates variable person. It can be any person. Similar is the case with the pronouns “he”, “it”, “they” etc., If pronouns are present in a sentence, then it cannot be considered as a statement in mathematics.
12. Consider the sentence:
There are 30 days in a month.
• On some occasions, this sentence is true.
For example, when the month being referred to, is April.
• On some occasions, this sentence is false.
For example, when the month being referred to, is August.
• There is ambiguity in this sentence. So it is not a statement.
13. Consider the sentence:
There are 40 days in a month.
• This sentence is always false. There is no ambiguity. So it is a statement.
14.  Consider the sentence:
(x+y)² = x² + y².
• On some occasions, this sentence is true.
   ♦ For example, when x = 1 and y = 0
• On some other occasions, this sentence is false.
   ♦ For example, when x = 2 and y = 3
• There is ambiguity in this sentence. So it is not a statement.
15. Consider the sentence:
There exists two real numbers x and y, such that (x+y)² = x² + y².  
• This sentence is true. There is no ambiguity. So it is a statement.

---

• Statements are denoted by small letters p, q, r etc.,
• For example, consider the statement:
For any two natural numbers x and y, the sum (x+y) is greater than zero.  
• We can write:
p: For any two natural numbers x and y, the sum (x+y) is greater than zero.

---

Let us see some solved examples:

Solved example 14.1
Check whether the following sentences are statements. Give reasons for your answer.
(i) 8 is less than 6.
(ii) Every set is a finite set.
(iii) The sun is a star.
(iv) Mathematics is fun.
(v) There is no rain without clouds.
(vi) How far is Chennai from here ? 
Solution:
Part (i):
8 is less than 6.
• This statement is false. There is no ambiguity. So it is a statement.
Part (ii):
Every set is a finite set.
• This statement is false. There is no ambiguity. So it is a statement.
(There are sets which are not finite)
Part (iii):
The sun is a star.
• This statement is true. There is no ambiguity. So it is a statement.
(It is scientifically proven that, sun is a star)
Part (iv):
Mathematics is fun.
• For a person who is interested in mathematics, the sentence is true.
• For a person who is not interested in mathematics, the sentence is false.
• There is ambiguity in this sentence. So it is not a statement.
Part (v):
There is no rain without clouds.
• This statement is true. There is no ambiguity. So it is a statement.
(It is scientifically proven that, clouds are required for rain to occur)
Part (vi):
How far is Chennai from here ?
• This is a question. Also, the word "here" is present. So it is not a statement.

---

From the above discussion and examples, the following three points can be noted:
1. We will be asked to say whether a given sentence is a statement or not.
2. The answer can be any one of the two given below:
Yes. The given sentence is a statement.
No. The given sentence is not a statement.
3. But besides answering “Yes” or “No”, we must write the reason also.

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The link to a few more solved examples is given below:

Exercise 14.1

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In the next section, we will see New statements from old.

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Chapter 13.18 - Miscellaneous Examples

In the previous section, we completed a discussion on the derivatives of trigonometric functions. In this section, we will see some miscellaneous examples.

Solved example 13.19
Find the derivative of f from the first principle, where f is given by
(i) $f(x) = \frac{2x+3}{x-2} $    (ii) $f(x) = x + \frac{1}{x}$
Solution:
Part (i):
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{2(x+h) + 3}{(x+h) - 2}~–~ \frac{2x+3}{x-2}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{2x + 2h + 3}{x+h - 2}~–~ \frac{2x+3}{x-2}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{(2x + 2h + 3)(x-2) ~-~(x+h-2)(2x+3)}{(x+h - 2)(x-2)}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{2x^2 - 4x + 2hx - 4h +3x - 6 - 2x^2 - 2hx + 4x - 3x - 3h +6}{(x+h - 2)(x-2)}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{  - 4h - 3h }{(x+h - 2)(x-2)}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{-7h}{h(x+h - 2)(x-2)} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{-7}{(x+h - 2)(x-2)} \right]}} &{} \\

{}&{}
& {~=~}& {\frac{-7}{(x - 2)(x-2)}} &{} \\

{}&{}
& {~=~}& {\frac{-7}{(x - 2)^2}} &{} \\

\end{array}$

Part (ii):
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[(x+h) + \frac{1}{(x+h)} \right]~-~\left[x + \frac{1}{x} \right]}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\frac{(x+h)^2 + 1}{(x+h)} \right]~-~\left[\frac{x^2 + 1}{x} \right]}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{\left[(x+h)^2 + 1 \right]x~-~\left[(x^2 + 1)(x+h) \right]}{(x+h)x}}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[(x+h)^2 + 1 \right]x~-~\left[(x^2 + 1)(x+h) \right]}{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[x(x+h)^2 + x \right]~-~\left[x^3 + x^2 h + x + h \right]}{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[x^3 + 2 x^2 h + h^2 x + x \right]~-~\left[x^3 + x^2 h + x + h \right]}{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x^3 + 2 x^2 h + h^2 x + x ~-~x^3 - x^2 h - x - h }{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x^2 h + h^2 x - h }{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h(x^2  + h x - 1) }{(x+h)xh} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x^2  + h x - 1 }{(x+h)x} \right]}} &{} \\

{}&{}
& {~=~}& {\frac{x^2  + (0) x - 1 }{(x+0)x}} &{} \\

{}&{}
& {~=~}& {\frac{x^2 - 1 }{x^2}} &{} \\

{}&{}
& {~=~}& {1 - \frac{1 }{x^2}} &{} \\

\end{array}$

Solved example 13.20
Find the derivative of f(x) from the first principle, where f(x) is
(i) sin x + cos x     (ii) x sin x
Solution:
Part (i):
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\sin (x+h) + \cos (x+h) \right]~-~\left[\sin x + \cos x \right]}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin (x+h) + \cos (x+h) - \sin x - \cos x}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin (x+h) + \cos (x+h) - \sin x - \cos x}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\sin (x+h) - \sin x \right]+ \left[\cos (x+h) - \cos x \right]}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[2 \cos \frac{2x+h}{2} \sin \frac{h}{2} \right]+ \left[-2 \sin \frac{2x+h}{2} \sin \frac{h}{2} \right]}{h} \right]}~\color{green}{\text{- - - I}}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2 \sin \frac{h}{2} \left[\cos \frac{2x+h}{2}~-~\sin \frac{2x+h}{2} \right]}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin \frac{h}{2}}{h/2} \right]}~ × ~\lim_{h\rightarrow 0}{\left[\cos \frac{2x+h}{2}~-~\sin \frac{2x+h}{2} \right]}} &{} \\

{}&{}
& {~=~}& {1~ × ~\left[\cos \frac{2x+0}{2}~-~\sin \frac{2x+0}{2} \right]} &{} \\

{}&{}
& {~=~}& {1~ × ~\left[\cos x~-~\sin x \right]} &{} \\

{}&{}
& {~=~}& {\cos x~-~\sin x} &{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Here we use the following identities:
    ♦ sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
    ♦ cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2

Part (ii):
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{(x+h) \sin (x+h) ~-~x \sin x}{h}   \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x \sin (x+h) + h \sin (x+h) - x \sin x}{h}\right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x\left[\sin (x+h)  -  \sin x \right] + h \sin (x+h)}{h}   \right]}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x\left[2 \cos \frac{2x+h}{2} \sin \frac{h}{2} \right] + h \sin (x+h)}{h} \right]}~\color{green}{\text{- - - I}}} &{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2x \cos \frac{2x+h}{2} \sin \frac{h}{2}}{h} \right]}~+~\lim_{h\rightarrow 0}{\left[\frac{h \sin (x+h)}{h} \right]}} &{} \\

{}&{}
& {~=~}& {\left[\lim_{h\rightarrow 0}{\left[x \cos \frac{2x+h}{2} \right]}~ × ~\lim_{h\rightarrow 0}{\left[\frac{\sin \frac{h}{2}}{h/2} \right]} \right]

~+~\lim_{h\rightarrow 0}{\left[ \sin (x+h) \right]}} &{} \\

{}&{}
& {~=~}& {\left[{\left[x \cos \frac{2x+0}{2} \right]}~ × ~1 \right]

~+~{\left[\sin (x+0) \right]}} &{} \\

{}&{}
& {~=~}& {x \cos x + \sin x} &{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Here we use the following identity:
    ♦ sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2

Solved example 13.21
Compute the derivative of
(i) f(x) = sin 2x  (ii) g(x) = cot x
Solution:
Part (i):
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {(\sin 2x)'} &{} \\

{}&{}
& {~=~}& {(2 \sin x \cos x)'} &{} \\

{}&{}
& {~=~}& {(2 \sin x)' (\cos x)~+~(2\sin x)(\cos x)' \color{green}{\text{- - - I}}} &{} \\

{}&{}
& {~=~}& {(2 \sin x)' (\cos x)~+~(2\sin x)(-\sin x)} &{} \\

{}&{}
& {~=~}& {(2 \sin x)' (\cos x)~-~2 \sin^2 x} &{} \\

{}&{}
& {~=~}& {\left[(2)' \sin x ~+~ (2) (\sin x)' \right]\cos x~-~2 \sin^2 x} &{} \\

{}&{}
& {~=~}& {\left[0 × \sin x ~+~ (2) (\cos x) \right]\cos x~-~2 \sin^2 x} &{} \\

{}&{}
& {~=~}& {\left[2 \cos x \right]\cos x~-~2 \sin^2 x} &{} \\

{}&{}
& {~=~}& {2 \cos^2 x~-~2 \sin^2 x} &{} \\

{}&{}
& {~=~}& {2 (\cos^2 x - \sin^2 x)} &{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Here we use the product rule: (uv)' = u'v + uv'

Part (ii):
• Here we want to find the derivative of cot x.
• This is already done before.
• See question number 11(v) of Exercise 13.2.

Solved example 13.22
Find the derivative of
(i) $\frac{x^5 - \cos x}{\sin x}$  (ii) $\frac{x + \cos x}{\tan x}$
Solution:
Part (i):
1. Here we can use quotient rule:
$\left(\frac{u}{v}\right)' = \frac{u' v - u v'}{u^2}$
• In our present case,
   ♦ u = x⁵ - cos x
   ♦ v = sin x
2. Thus we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\frac{\left[(x^5 - \cos x)' \sin x \right] ~-~\left[(x^5 - \cos x) (\sin x)' \right]}{\sin^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{\left[(5 x^4 + \sin x) \sin x \right] ~-~\left[(x^5 - \cos x) (\cos x) \right]}{\sin^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{\left[5 x^4 \sin x + \sin^2 x \right] ~-~\left[x^5 \cos x - \cos^2 x \right]}{\sin^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{5 x^4 \sin x + \sin^2 x  - x^5 \cos x + \cos^2 x}{\sin^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{5 x^4 \sin x - x^5 \cos x + 1}{\sin^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{- x^5 \cos x + 5 x^4 \sin x + 1}{\sin^2 x}} &{} \\

\end{array}$

Part (ii):
1. Here we can use quotient rule:
$\left(\frac{u}{v}\right)' = \frac{u' v - u v'}{u^2}$
• In our present case,
   ♦ u = x + cos x
   ♦ v = tan x
2. Thus we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\frac{\left[(x + \cos x)' \tan x \right] ~-~\left[(x + \cos x) (\tan x)' \right]}{\tan^2 x}} &{} \\

{}&{}
& {~=~}& {\frac{\left[(1- \sin x) \tan x \right] ~-~\left[(x + \cos x) (\sec^2 x) \right]}{\tan^2 x}} &{} \\

\end{array}$

---

Link to a few more miscellaneous examples is given below:

Miscellaneous Exercise on chapter 13

---

In the next chapter, we will see mathematical reasoning.

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Chapter 13.17 - Derivatives of Trigonometric Functions

In the previous section, we saw the derivatives of polynomial functions. In this section, we will see the derivatives of trigonometric functions.

• We have already seen the derivative of sin x.
    ♦ That is., when f(x) = sin x, f'(x) will be cos x
• We obtained this result using first principle.
    ♦ That is., we used the formula:
$f'(x) = \lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}$
[See fig.13.33 in section 13.13]
• To find derivatives of various trigonometric functions, we need to know the derivative of cos x also.
• To find the derivative of cos x, we must use the first principle.
• Once we know the derivative of both sin x and cos x, we can use sum rule, product rule, quotient rule etc.,

• The derivative of cos x can be obtained as follows:
$\begin{array}{ll} {}&{f'(x)} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\

  {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\cos (x+h)] – [cos x]}{h} \right]}} &{} \\

  {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\cos x \cos h ~-~\sin x \sin h] – [cos x]}{h} \right]}} &{} \\

  {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\cos x (\cos h - 1)~-~ \sin x \sin h}{h} \right]}} &{} \\

  {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\cos x (\cos h - 1)}{h}
\right]}~-~\lim_{h\rightarrow 0}{\left[\frac{\sin x \sin h}{h} \right]}} &{} \\

  {}&{} & {~=~}& {\cos x \lim_{h\rightarrow 0}{\left[\frac{\cos h - 1}{h}
\right]}~-~\sin x \lim_{h\rightarrow 0}{\left[\frac{\sin h}{h} \right]}} &{} \\

{}&{} & {~=~}& {\cos x  × 0~-~\sin x  × 1} &{} \\
  {}&{} & {~=~}& {- \sin x} &{} \\ 

\end{array}$

• Now we have two useful results:
    ♦ If f(x) = sin x, then f'(x) = cos x  
    ♦ If f(x) = cos x, then f'(x) = -sin x

• We have already seen the graph of sin x and it's derivative. Now we will see the graph of cos x and it's derivative. It can be written in 5 steps:

1. We know that, f’(x) is a function. So we must  be able to plot f’(x).
2. In fig.13.39 below, both f(x) and f’(x) are plotted in the same graph.

Fig.13.39

3. The two graphs are plotted in different colors.
    ♦ The red curve represents f(x)
    ♦ The green curve represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at $\left(x=\frac{\pi}{6}\right)$. That is., we want $f' \left(\frac{\pi}{6}\right)$
• For that, we mark point A on the red curve. Here, $\left(x=\frac{\pi}{6}\right)$
• Next we draw a vertical line through A. This vertical line meets the green curve at A'.
• $f' \left(\frac{\pi}{6}\right)$ will be equal to the y-coordinate of A'. In this case, it is $-\frac{1}{2}$
• We can verify this theoretically:
$f' \left(\frac{\pi}{6}\right)~=~- \sin \left(\frac{\pi}{6}\right)~=~- \frac{1}{2}$
(ii) Suppose that, we want the derivative of f(x) at $\left(x = -\pi \right)$. That is., we want $f' \left(-\pi \right)$
• For that, we mark point B on the red curve. Here, $\left(x = -\pi \right)$
• Next we draw a vertical line through B. This vertical line meets the green curve at B'.
• $f' \left(-\pi \right)$ will be equal to the y-coordinate of B'. In this case, it is 0.
• We can verify this theoretically:
$f' \left(-\pi \right)~=~- \sin \left(-\pi \right)~=~0$
5. Further more, the reader may verify the tangents also:
• Through the point A, draw a line at a slope of $- \frac{1}{2}$. This line will be tangential to f(x).
• Through the point B, draw a line at a slope of 0 (a horizontal line). This line will be tangential to f(x).   

---

Let us see some solved examples:

Solved example 13.17
Compute the derivative of tan x
Solution:
1. We can apply the quotient rule:
$\left(\frac{u}{v} \right)' = \frac{u'v - uv' }{v^2}$
2. In our present case:
u = sin x. So u' = cos x
v = cos x. So v' = -sin x
3. Thus we get:

$\begin{array}{ll}
{}&{\left(\frac{\sin x}{\cos x} \right)'}
& {~=~}& {\left(\frac{u}{v} \right)'}
&{} \\

{}&{}
& {~=~}& {\frac{[\cos x × \cos x] - [\sin x × -\sin x] }{\cos^2 x}}
&{} \\

{}&{}
& {~=~}& {\frac{\cos^2 x + \sin^2 x}{\cos^2 x}}
&{} \\

{}&{}
& {~=~}& {\frac{1}{\cos^2 x}}
&{} \\

{}&{}
& {~=~}& {\sec^2 x}
&{} \\

\end{array}$

Alternate method:
We can use first principle:

$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\tan(x+h) – \tan x}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{ \frac{\sin (x+h)}{\cos (x+h)} - \frac{\sin x}{\cos x}}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{ \frac{\sin (x+h) \cos x~-~\cos (x+h) \sin x}{\cos (x+h) \cos x}}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin (x+h) \cos x~-~\cos (x+h) \sin x}{h\;\cos (x+h) \cos x} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin (x+h - x)}{h\;\cos (x+h) \cos x} \right]}~\color{green}{\text{- - - (I)}}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin h}{h\;\cos (x+h) \cos x} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin h}{h} \right]}~ × ~\lim_{h\rightarrow 0}{\left[\frac{1}{\cos (x+h) \cos x} \right]}}
&{} \\

{}&{}
& {~=~}& {1~ × ~\frac{1}{\cos^2 x}}
&{} \\

{}&{}
& {~=~}& {\sec^2 x}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Here we use the identity:
sin (A-B) = sin A cos B - cos A sin B

---

A graphical description can be written in 5 steps:
1. We know that, f’(x) is a function. So we must  be able to plot f’(x).
2. In fig.13.40 below, both f(x) and f’(x) are plotted in the same graph.

Fig.13.40

3. The two graphs are plotted in different colors.
    ♦ The red curve represents f(x)
    ♦ The green curve represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at $\left(x=\frac{\pi}{3}\right)$. That is., we want $f' \left(\frac{\pi}{3}\right)$
• For that, we mark point A on the red curve. Here, $\left(x=\frac{\pi}{3}\right)$
• Next we draw a vertical line through A. This vertical line meets the green curve at A'.
• $f' \left(\frac{\pi}{3}\right)$ will be equal to the y-coordinate of A'. In this case, it is 4
• We can verify this theoretically:
$f' \left(\frac{\pi}{3}\right)~=~ \sec^2 \left(\frac{\pi}{3}\right)~=~\frac{1}{\cos^2 \left(\frac{\pi}{3}\right)}~=~\frac{1}{(1/2)^2}~=~4$
(ii) Suppose that, we want the derivative of f(x) at $\left(x = -\pi \right)$. That is., we want $f' \left(-\pi \right)$
• For that, we mark point B on the red curve. Here, $\left(x = -\pi \right)$
• Next we draw a vertical line through B. This vertical line meets the green curve at B'.
• $f' \left(-\pi \right)$ will be equal to the y-coordinate of B'. In this case, it is 1.
• We can verify this theoretically:
$f' \left(-\pi\right)~=~ \sec^2 \left(-\pi\right)~=~\frac{1}{\cos^2 \left(-\pi\right)}~=~\frac{1}{(-1)^2}~=~1$
5. Further more, the reader may verify the tangents also:
• Through the point A, draw a line at a slope of 4. This line will be tangential to f(x).
• Through the point B, draw a line at a slope of 1. This line will be tangential to f(x).

Solved example 13.18
Compute the derivative of f(x) = sin² x
Solution:
1. We can write sin² x as sin x . sin x
Now we can apply the product rule: (uv)' = u'v + uv'
2. The steps are shown below:
$\begin{array}{ll}
{}&{\left(\sin^2 x \right)'}
& {~=~}& {\left(\sin x \, . \, \sin x \right)'}
&{} \\

{}&{}
& {~=~}& {(\sin x)' × \sin x~+~\sin x  × (\sin x)'}
&{} \\

{}&{}
& {~=~}& {\cos x × \sin x~+~\sin x  × \cos x}
&{} \\

{}&{}
& {~=~}& {2 \sin x \cos x}
&{} \\

{}&{}
& {~=~}& {\sin 2x}
&{} \\

\end{array}$

A graphical description can be written in 5 steps:
1. We know that, f’(x) is a function. So we must  be able to plot f’(x).
2. In fig.13.41 below, both f(x) and f’(x) are plotted in the same graph.

Fig.13.41

3. The two graphs are plotted in different colors.
    ♦ The red curve represents f(x)
    ♦ The green curve represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at $\left(x=\frac{\pi}{6}\right)$. That is., we want $f' \left(\frac{\pi}{6}\right)$
• For that, we mark point A on the red curve. Here, $\left(x=\frac{\pi}{6}\right)$
• Next we draw a vertical line through A. This vertical line meets the green curve at A'.
• $f' \left(\frac{\pi}{6}\right)$ will be equal to the y-coordinate of A'. In this case, it is 4
• We can verify this theoretically:
$f' \left(\frac{\pi}{6}\right)~=~ \sin \left(\frac{2 \pi}{6}\right)~=~ \sin \left(\frac{\pi}{3}\right) ~=~\frac{\sqrt3}{2}$
(ii) Suppose that, we want the derivative of f(x) at $\left(x=-\frac{\pi}{4}\right)$. That is., we want $f' \left(-\frac{\pi}{4}\right)$
• For that, we mark point B on the red curve. Here, $\left(x=-\frac{\pi}{4}\right)$
• Next we draw a vertical line through B. This vertical line meets the green curve at B'.
• $f' \left(-\frac{\pi}{4}\right)$ will be equal to the y-coordinate of B'. In this case, it is 4
• We can verify this theoretically:
$f' \left(-\frac{\pi}{4}\right)~=~ \sin \left(\frac{-2 \pi}{4}\right)~=~ \sin \left(-\frac{\pi}{2}\right)~=~ -\sin \left(\frac{\pi}{2}\right) ~=~-1$
5. Further more, the reader may verify the tangents also:
• Through the point A, draw a line at a slope of $\frac{\sqrt3}{2}$. This line will be tangential to f(x).
• Through the point B, draw a line at a slope of -1. This line will be tangential to f(x).

---

• We have drawn a large number of graphs. Those graphs were drawn to demonstrate the close relation between f(x) and f'(x).
• From now on wards, we will not draw any more graphs. We will concentrate only on the method for obtaining f'(x).
• However, the interested reader may continue to draw them. 

---

Link to a few more solved examples is given below:

Exercise 13.2

---

In the next section, we will see some miscellaneous examples.

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Chapter 13.16 - Derivatives of Polynomial Functions

In the previous section, we saw the algebra of derivatives. We also saw the derivative f'(x) when f(x) = x². In this section, we will see the derivative when the power of x is any +ve integer. Later in this section, we will see derivatives of polynomial functions also.

Derivative when the power of x is any positive integer

It can be written in 4 steps:
1. We want f'(x) where f(x) = xⁿ
2. We can write:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{(x+h)^n – x^n}{h} \right]}}
&{} \\

\end{array}$

3. So first we need to find $\frac{(x+h)^n – x^n}{h}$
• It can be calculated as follows:

$\begin{array}{ll}
{}&{(x+h)^n}
& {~=~}& {\rm{\left(nC_0 \right)}x^n~+~\rm{\left(nC_1 \right)}x^{n-1} h~+~\rm{\left(nC_2 \right)}x^{n-2} h^2~+~\rm{\left(nC_3 \right)}x^{n-3} h^3~+~.~.~.~+~\rm{\left(nC_n \right)}h^n ~ \color {green} {\text{- - - (I)}}}
&{} \\

{\Rightarrow}&{(x+h)^n}
& {~=~}& {x^n~+~\rm{n}x^{n-1} h~+~\rm{\left(nC_2 \right)}x^{n-2} h^2~+~\rm{\left(nC_3 \right)}x^{n-3} h^3~+~.~.~.~+~\rm{n}h^n}
&{} \\

{\Rightarrow}&{(x+h)^n~-~x^n}
& {~=~}& {\rm{n}x^{n-1} h~+~\rm{\left(nC_2 \right)}x^{n-2} h^2~+~\rm{\left(nC_3 \right)}x^{n-3} h^3~+~.~.~.~+~\rm{n}h^n}
&{} \\

{\Rightarrow}&{(x+h)^n~-~x^n}
& {~=~}& {h\left[\rm{n}x^{n-1}~+~\rm{\left(nC_2 \right)}x^{n-2} h^1~+~\rm{\left(nC_3 \right)}x^{n-3} h^2~+~.~.~.~+~\rm{n}h^{n-1}\right]}
&{} \\

{\Rightarrow}&{\frac{(x+h)^n~-~x^n}{h}}
& {~=~}& {\rm{n}x^{n-1}~+~\rm{\left(nC_2 \right)}x^{n-2} h^1~+~\rm{\left(nC_3 \right)}x^{n-3} h^2~+~.~.~.~+~\rm{n}h^{n-1}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we apply the binomial theorem.

4. Now the calculations in (2) can be completed:

$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{(x+h)^n – x^n}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\rm{n}x^{n-1}~+~\rm{\left(nC_2 \right)}x^{n-2} h^1~+~\rm{\left(nC_3 \right)}x^{n-3} h^2~+~.~.~.~+~\rm{n}h^{n-1} \right]}}
&{} \\

{}&{}
& {~=~}& {{n x^{n-1}} ~ \color {green} {\text{- - - (I)}}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
When the limit is applied, all terms with 'h' will become zero.

---

This formula can be proved using mathematical induction also. It can be written in 5 steps:
1. We want to prove the statement:
If f(x) = xⁿ, then f'(x) = nx^n-1.
2. First we must prove the statement for n = 1
   ♦ If n = 1, then f(x) = x¹ = x
   ♦ Then f'(x) = 1 × x^1-1 = 1 × x⁰ = 1
• So the statement is true for n = 1.
3. Next we assume that the statement is true for n = k
   ♦ If n = k, then f(x) = x^k
   ♦ Then f'(x) = k × x^k-1
• That is., we assume that, f'(x^k) = k × x^k-1 is true.
4. Next we must prove that, the statement is true for n = k+1
   ♦ If n = k+1, then f(x) = x^k+1
   ♦ Then f'(x) = (k+1) × x^(k+1)-1 = (k+1) × x^k .
• That is.,we must prove that,
f'(x^k+1) = (k+1) × x^(k+1)-1 = (k+1) × x^k is true.
5. This can be done as follows:

$\begin{array}{ll}
{}&{f'(x^{k+1})}
& {~=~}& {(k+1)x^k}
&{} \\

{\Rightarrow}&{f'(x^k \; .\; x)}
& {~=~}& {(k+1)x^k}
&{} \\

{\Rightarrow}&{f'(x^k) \; .\; x ~+~x^k \; .\; f'(x)}
& {~=~}& {(k+1)x^k ~~ \color {green} {\text{- - - (I)}}}
&{} \\

{\Rightarrow}&{f'(x^k) \; .\; x ~+~x^k \; .\; 1}
& {~=~}& {(k+1)x^k ~~ \color {green} {\text{- - - (II)}}}
&{} \\

{\Rightarrow}&{k\; .\; x^{k-1} \; .\; x ~+~x^k \; .\; 1}
& {~=~}& {(k+1)x^k ~~ \color {green} {\text{- - - (III)}}}
&{} \\

{\Rightarrow}&{k\; .\; x^{k}~+~x^k}
& {~=~}& {(k+1)x^k}
&{} \\

{\Rightarrow}&{(k+1)x^k}
& {~=~}& {(k+1)x^k}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Here we use the product rule on the LHS
• Line marked as II:
Here we use the fact that:
If f(x) = x, then f'(x) = 1
• Line marked as III:
Here we use the result in (3)

◼ Thus we have proved the statement in (4). And so, the statement in (1) is proved.

---

• So now we have a useful formula:
If f(x) = xⁿ, then f'(x) = nx^n-1.
   ♦ We proved this when n is a +ve integer.
   ♦ In fact, this formula is valid when n is any real number. We will see it's proof in higher classes.
• This formula is known as Theorem 6.

---

Let us see an example.
If f(x) = x⁹, then f'(x) will be 9x⁸

◼ A graphical description can be written in 5 steps:
1. We know that, f’(x) is a function. So we must  be able to plot f’(x).
2. In fig.13.38 below, both f(x) and f’(x) are plotted in the same graph.

Fig.13.38

3. The two graphs are plotted in different colors.
    ♦ The red curve represents f(x)
    ♦ The green curve represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at (x = 0.75). That is., we want f’(0.75)
• For that, we mark point A on the red curve. Here, (x = 0.75)
• Next we draw a vertical line through A. This vertical line meets the green curve at A'.
• f’(0.75) will be equal to the y-coordinate of A'. In this case, it is 0.9.
• We can verify this theoretically:
f'(0.75) = 9 × (0.75)⁸ = 0.9010
(ii) Suppose that, we want the derivative of f(x) at (x = -0.8). That is., we want f’(-0.8)
• For that, we mark point B on the red curve. Here, (x = -0.8)
• Next we draw a vertical line through B. This vertical line meets the green curve at B'.
• f’(-0.8) will be equal to the y-coordinate of B'. In this case, it is 1.51.
• We can verify this theoretically:
f'(-0.8) = 9 × (-0.8)⁸ = 1.509
5. Further more, the reader may verify the tangents also:
• Through the point A, draw a line at a slope of 0.9. This line will be tangential to f(x).
• Through the point B, draw a line at a slope of 1.51. This line will be tangential to f(x).

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Derivative of polynomial functions

This can be written in 5 steps:
1. Consider the polynomial function:
f(x) = a_nxⁿ + a_n-1 x^n-1 + a_n-2 x^n-2 + . . . + a₁x¹ + a₀ x⁰.
   ♦ a₀, a₁, a₂ etc., are real numbers.
• We want to find f'(x)
2. We can consider each term as a separate function.
• Then we can find the derivative of each function using theorem 6 that we saw above.
• For example, the derivative of a_n-1 x^n-1 will be a_n-1 (n-1)x^n-2.
3. After finding each derivative, we can apply the sum rule.
4. So we get:
f'(x) = a_n nx^n-1 + a_n-1 (n-1)x^n-2 + a_n-2 (n-2)x^n-3 + . . . + a₁(1)x⁰ + 0.
5. This method for finding the derivative of polynomial functions is known as theorem 7.

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Let us see some solved examples:

Solved example 13.14
Compute the derivative of 6x¹⁰⁰ - x⁵⁵ + x
Solution:
1. Derivative of the first term is:
6 × 100x⁹⁹ = 600x⁹⁹.
2. Derivative of the second term is:
- 55x⁵⁴.
3. Derivative of the third term is: 1.
4. Applying the sum rule, we get:
f'(x) = 600x⁹⁹ - 55x⁵⁴ + 1

Solved example 13.15
Compute the derivative of
1 + x + x² + x³ + . . . + x⁵⁰ at x = 1
Solution:
1. We see that, there are 51 terms in f(x).
2. The derivative f'(x) is:
0 + 1 + 2x + 3x² + 4x³ + . . . + 50x⁴⁹
• So there are 50 terms in f'(x)
3. Substituting (x=1) in f'(x), we get:
f'(1) = 1 + 2 + 3 + 4 + . . . + 50
• This is in the form of an A.P where,
   ♦ First term = 1
   ♦ Last term = 50
   ♦ Number of terms (n) = 50
4. So sum of the A.P can be obtained as:
$S~=~\frac{n(\text{First term + Last term})}{2}~=~\frac{50(1+50)}{2}$ = 25 × 51 = 1275

Solved example 13.16
Find the derivative of $f(x)=\frac{x+1}{x}$
Solution:
1. We can apply the quotient rule:
$\left(\frac{u}{v} \right)' = \frac{u'v - uv' }{v^2}$
2. In our present case:
    ♦ u = (x+1). So u' = 1
    ♦ v = x. So v' = 1
3. Thus we get:
$\left(\frac{u}{v} \right)' = \frac{1(x) - (x+1)1 }{x^2} = \frac{x - x - 1}{x^2} = - \frac{1}{x^2}$

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In the next section, we will see derivatives of trigonometric functions.

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Chapter 13.15 - Algebra of Derivatives

In the previous section, we saw the graphical representation of the derivative. We saw some solved examples also. In this section, we will see algebra of derivatives.

• We obtained derivative by using the principles of limits.
So we can expect:
   ♦ algebra of derivatives
   ♦ to be similar to
   ♦ algebra of limits.
• For example, let us consider the sum of two functions. We know that:
   ♦ limit of the sum [f(x) + g(x)]
         [This sum is written as: (f+g)(x)]
   ♦ is equal to
   ♦ limit of f(x) + limit of g(x)
• We can extend this to derivatives:
   ♦ derivative of the sum (f+g)(x)
   ♦ is equal to
   ♦ derivative of f(x) + derivative of g(x)
• That is, (f+g)'(x) = f'(x) + g'(x)
• The proof can be written in 10 steps:
1. Suppose that, x₁ is an allowable input x-value for the two functions f and g
• First, calculate f(x₁)
• Next, calculate g(x₁)
• Next, calculate the sum [f(x₁) + g(x₁)]
• Finally, calculate (f+g)(x₁). It will be same as the above sum.
2. This can be illustrated using an example:
   ♦ Let f(x) = x² + 1
   ♦ Let g(x) = 2x + 3
• Then (f+g)(x) = x² + 2x + 4
3. Let us try an input x-value of ‘2’
   ♦ f(2) = 2² + 1 = 5
   ♦ g(2) = 2 × 2 + 3 = 7
   ♦ f(2) + g(2) = 5 + 7 = 12
4. Let us try the same input x-value for (f+g)(x). We get:
(f+g)(2) = (2² + 2 × 2 + 4) = (4 + 4 + 4) = 12
5. By comparing the results in (3) and (4), we see that:
The statement in (1) is true.
6. Now we revisit the familiar fig.13.31 in section 13.13. Recall that, we used that fig.13.31 to obtain the derivative. We are going to use a modified version of that fig.13.31. It is shown below:

Fig.13.37

7. This time, the red curve represents the graph of (f+g)x.
   ♦ So the coordinates of P will be: (x,(f+g)(x))
   ♦ Also, the coordinates of Q will be: ((x+h), (f+g)(x+h))
8. Using the coordinates, we get:
   ♦ Altitude QR = [(f+g)(x+h) – (f+g)(x)]
   ♦ Base PR = [(x+h) – x] = h
9. So the slope of the yellow line = $\frac{(f+g)(x+h) – (f+g)(x)}{h}$
10. So slope of the tangent at P will be:
(f+g)'(x) =$\lim_{h\rightarrow 0}{\left[\frac{(f+g)(x+h) – (f+g)(x)}{h} \right]}$
• This can be simplified as follows:

$\begin{array}{ll}
{}&{(f+g)'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{(f+g)(x+h) – (f+g)(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[f(x+h) + g(x+h)] – [f(x) + g(x)]}{h} \right]~\color{green}{\text{- - - I}}}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[f(x+h) - f(x) ] + [g(x+h) - g(x)]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[f(x+h) - f(x) ]}{h}~+~\frac{[g(x+h) - g(x)]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[f(x+h) - f(x) ]}{h}\right]}~+~\lim_{h\rightarrow 0}{\left[\frac{[g(x+h) - g(x)]}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {f'(x)~+~g'(x)}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we apply the result in (5).

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• Using the same technique, we will be able to prove that:
(f-g)'(x) = f'(x) - g'(x)

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• For multiplication, we use the formula:
(f.g)'(x) = f'(x).g(x) + f(x).g'(x)
• We will see the proof in higher classes.
• This formula is easier to remember, if we use the following substitutions:
   ♦ f(x) = u and f'(x) = u'
   ♦ g(x) = v and g'(x) = v'
   ♦ (f.g)'(x) = (uv)'
• Now the formula becomes: (uv)' = u'v + uv'

• This is known as Leibnitz rule for differentiating product of functions. 

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• For division, we use the formula:
$\left(\frac{f}{g} \right)'(x)~=~ \frac{f'(x).g(x) - f(x).g'(x)}{\left(g(x) \right)^2}$
• This formula is easier to remember, if we use the same substitutions as for multiplication. Then the formula becomes:
$\left(\frac{u}{v} \right)'(x)~=~ \frac{u'.v - u.v'}{v^2}$

---

• Now we have the formulas for addition, subtraction, multiplication and division.
• The four formulas are together known as theorem 5.

---

Using theorem 5, we can compute some common derivatives.
◼ First we will find f’(x) when f(x) = x

$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x+h - x}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h}{h} \right]}}
&{} \\

{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[1 \right]}}
&{} \\

{}&{}
& {~=~}& {1}
&{} \\

\end{array}$

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Now we will calculate f'(x) when f(x) = 10x

$\begin{array}{ll}
{}&{f'(10x)}
& {~=~}& {f'(x) + f'(x) + f'(x)~ + ~.~.~.~+f'(x)~\text{(10 terms)}~\color{green}{\text{- - - I}}}
&{} \\

{}&{}
& {~=~}& {1 + 1 + 1~+ ~.~.~.~+~1~\text{(10 terms)}}
&{} \\

{}&{}
& {~=~}& {10}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we apply the sum rule.

---

• The same result can be obtained by applying product rule also. For that, we write:
   ♦ f(x) = u = 10
   ♦ g(x) = v = x
   ♦ f(x).g(x) = uv = 10x
• Applying the product rule, we get:

$\begin{array}{ll}
{}&{(uv)'}
& {~=~}& {u' v + u v'}
&{} \\

{}&{}
& {~=~}& {f'(10) × x + 10 × g'(x)}
&{} \\

{}&{}
& {~=~}& {0 × x + 10 × 1~\color{green}{\text{- - - I}}}
&{} \\

{}&{}
& {~=~}& {10}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we apply the following facts:
   ♦ Derivative of a constant function is zero.
   ♦ Derivative of x is 1.

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• In the same way, we can obtain the derivative of f(x) = x². For that, we write:
   ♦ f(x) = u = x
   ♦ g(x) = v = x
   ♦ f(x).g(x) = uv = x²
• Applying the product rule, we get:

$\begin{array}{ll}
{}&{(uv)'}
& {~=~}& {u' v + u v'}
&{} \\

{}&{}
& {~=~}& {f'(x) × x + x × g'(x)}
&{} \\

{}&{}
& {~=~}& {1 × x + x × 1~\color{green}{\text{- - - I}}}
&{} \\

{}&{}
& {~=~}& {x + x}
&{} \\

{}&{}
& {~=~}& {2x}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we apply the following fact:
   ♦ Derivative of x is 1.

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Now we know the derivative f'(x) when f(x) = x². In the next section, we will see the derivative when the power of x is any +ve integer.

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