21.13 - Properties of Logarithm

In the previous section, we saw some basic details about logarithmic functions. In this section, we will see some of their properties.

Property I: $\rm{\log_a p \,=\, \frac{\log_b p}{\log_b a}}$
   ♦ On the L.H.S, we have log of p to the base a.
   ♦ On the R.H.S, we have log of p to the base b.
• So this property gives us the method to connect two logs of the same number when the bases are different.
• This property is also known as change of base rule.
• We can write the proof in 5 steps:

1. Let $\rm{\log_a p = \alpha}$
Then $\rm{a^{\alpha} = p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b a = \gamma}$
Then $\rm{b^{\gamma} = a}$
4. Substituting (3) in (1), we get:
$\rm{\left(b^{\gamma} \right)^{\alpha} = p}$
⇒ $\rm{b^{\gamma \alpha} = p}$
5. Substituting (4) in (2), we get:
$\rm{b^{\beta} = p = b^{\gamma \alpha}}$
⇒ $\rm{\beta = \gamma \alpha}$
⇒ $\rm{\alpha = \frac{\beta}{\gamma}}$
⇒ $\rm{\log_a p = \frac{\log_b p}{\log_b a}}$

Property II: $\rm{\log_b pq \,=\, \log_b p + \log_b q}$
   ♦ On the L.H.S, we have log of a product pq.
   ♦ On the R.H.S, we have addition of two individual logs.
• So this property helps us to simplify logarithmic equations when products are involved.
• We can write the proof in 4 steps:
1. Let $\rm{\log_b pq = \alpha}$
Then $\rm{b^{\alpha} = pq}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b q = \gamma}$
Then $\rm{b^{\gamma} = q}$
4. Substituting (2) and (3) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\gamma}$
⇒ $\rm{b^\alpha = b^{\beta + \gamma}}$
⇒ $\rm{\alpha = \beta + \gamma}$
⇒ $\rm{\log_b pq = \log_b p + \log_b q}$

Property III: $\rm{\log_b \left(\frac{p}{q} \right) \,=\, \log_b p - \log_b q}$
   ♦ On the L.H.S, we have log of a ratio $\frac{p}{q}$.
   ♦ On the R.H.S, we have subtraction of one individual log from another.
• So this property helps us to simplify logarithmic equations when ratios are involved.
• We can write the proof in 4 steps:
1. Let $\rm{\log_b \left(\frac{p}{q} \right) = \alpha}$
Then $\rm{b^{\alpha} = \frac{p}{q}}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b q = \gamma}$
Then $\rm{b^{\gamma} = q}$
4. Substituting (2) and (3) in (1), we get:
$\rm{b^\alpha = \frac{b^\beta}{b^\gamma}}$
⇒ $\rm{b^\alpha = b^{\beta - \gamma}}$
⇒ $\rm{\alpha = \beta - \gamma}$
⇒ $\rm{\log_b pq = \log_b p - \log_b q}$

Property IV: $\rm{\log_b p^2 \,=\, 2 \log_b p}$
   ♦ On the L.H.S, we have log of a square.
   ♦ On the R.H.S, we have individual log multiplied by 2.
• So this property helps us to simplify logarithmic equations when squares are involved.
• We can write the proof in 3 steps:
1. Let $\rm{\log_b (p \times p) = \alpha}$
Then $\rm{b^{\alpha} = p \times p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Substituting (2) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\beta}$
⇒ $\rm{b^\alpha = b^{\beta + \beta}}$
⇒ $\rm{\alpha = \beta + \beta}$
⇒ $\rm{\log_b (p \times p) = \log_b p + \log_b p}$
⇒ $\rm{\log_b p^2 = 2 \log_b p}$

Property V: $\rm{\log_b p^3 \,=\, 3 \log_b p}$
   ♦ On the L.H.S, we have log of a cube.
   ♦ On the R.H.S, we have individual log multiplied by 3.
• So this property helps us to simplify logarithmic equations when cubes are involved.
• We can write the proof in 3 steps:
1. Let $\rm{\log_b (p \times p \times p) = \alpha}$
Then $\rm{b^{\alpha} = p \times p \times p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Substituting (2) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\beta \times b^\beta}$
⇒ $\rm{b^\alpha = b^{\beta + \beta + \beta}}$
⇒ $\rm{\alpha = \beta + \beta + \beta}$
⇒ $\rm{\log_b (p \times p \times p) = \log_b p + \log_b p + \log_b p}$
⇒ $\rm{\log_b p^3 = 3 \log_b p}$

Property VI: $\rm{\log_b p^n \,=\, n \log_b p}$
   ♦ On the L.H.S, we have log of a n^th power.
   ♦ On the R.H.S, we have individual log multiplied by n.
• So this property helps us to simplify logarithmic equations when n^th power is involved.
• The proof can be written using the principles of mathematical induction.

Property VII: $\rm{\log_p p \,=\, 1}$
   ♦ On the L.H.S, we have same number and base.
   ♦ On the R.H.S, we have 1.
We can write the proof as follows:
• Let $\rm{\log_p p = \alpha}$
• Then $\rm{p^{\alpha} = p}$
⇒ $\rm{p^{\alpha} = p^1}$
⇒ $\rm{\alpha = 1}$

Property VIII: $\rm{\log_b p \,=\, \frac{1}{\log_p b}}$
   ♦ On the L.H.S, we have base b and number p.
   ♦ On the R.H.S, we have base p and number b.
   ♦ So base and number are interchanged.
We can write the proof as follows:
• Applying the change of base rule we get:
$\rm{\log_b p \,=\, \frac{\log_p p}{\log_p b} \,=\, \frac{1}{\log_p b}}$ 

Property IX: $\rm{\log_p {p^x} \,=\, x}$
We can write the proof as follows:
• Let $\rm{\log_p p^x = \alpha}$
• Then $\rm{p^{\alpha} = p^x}$
⇒ $\rm{\alpha = x}$

• Note that, this property is used when we first begin to learn about logarithms. For example:
$\rm{\log_5 625 = \log_5 5^4 = 4}$

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Now we will see some solved examples

Solved example 21.43
Express as a single logarithm:
$\rm{2 \log x - 5 \log y + 3 \log z}$
Solution:
1. 2 log x = log x².
2. −5 log y = −log y⁵.
3. 3 log z = log z³.
4. log x² − log y⁵ + log z³ = $\rm{\frac{x^2 z^3}{y^5}}$

Solved example 21.44
Expand using the properties of logarithm:
$\rm{\log_5 \sqrt[3]{x}}$
Solution:
$\rm{\log_5 \sqrt[3]{x}}$
= $\rm{\log_5 \left(x^{\frac{1}{3}} \right)}$ 
= $\rm{\frac{1}{3} \log_5 x}$

Solved example 21.45
Evaluate log₅ 28
Solution:
1. We could find the solution easily if it is "25" instead of "28".
2. We cannot directly use the calculator because base is 5.
3. So we will apply the change of base rule. We get:
$\rm{\log_5 28 \,=\,\frac{\log_{10} 28}{\log_{10} 5}\,=\,\frac{1.44715}{0.69897}\,=\,2.07040}$

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In the next section, we will see exponential and logarithmic equations.

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21.12 - Logarithmic Functions

In the previous section, we completed a discussion on exponential functions. In this section, we will see logarithmic functions.

Some basic details about logarithmic functions can be written in 3 steps:
1. Consider the exponential function that we saw in the previous section: y = f(x) = b^x.
♦ where b > 0 and b ≠ 1.
2. Here "b" is a constant. x and y are variables.
• If we give an input x, we will get an output y.
• For example, if b = 2 and x = 3, we get: y = 2³ = 8.
• In such a situation, we say that:
Logarithm of 8 to the base 2 is 3
• In general, if y = b^x, we say that:
Logarithm of y to the base b is x.
3. Logarithm of y to the base b is denoted as log_b y.
• So if y = b^x, we can write: log_b y = x.

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Let us see some solved examples:

Solved example 21.41
Evaluate the following logarithms:
(a) log₁₀ 1000 (b) log₄ 16 (c) log₅ 625 (d) log_1/6 36 (e) $\log_9 \frac{1}{531441}$ (f) $\log_{\frac{3}{2}} \frac{27}{8}$
Solution:
Part (a)
1. Let log₁₀ 1000 = x
2. Then we can write: 1000 = 10^x.
3. We have: 1000 = 10³.
4. From (2) and (3), we get: 10^x = 10³.
• In the above equation, bases are the same. So equating the powers, we get: x = 3

Part (b)
1. Let log₄ 16 = x
2. Then we can write: 16 = 4^x.
3. We have: 16 = 4².
4. From (2) and (3), we get: 4^x = 4².
• In the above equation, bases are the same. So equating the powers, we get: x = 2

Part (c)
1. Let log₅ 625 = x
2. Then we can write: 625 = 5^x.
3. We have: 625 = 5⁴.
4. From (2) and (3), we get: 5^x = 5⁴.
• In the above equation, bases are the same. So equating the powers, we get: x = 4

Part (d)
1. Let $\log_{\frac{1}{6}} 36$ = x
2. Then we can write: $36 = \left(\frac{1}{6} \right)^x$.
⇒ $36 = 6^{-x}$
3. We have: 36 = 6².
4. From (2) and (3), we get: 6^−x = 6².
• In the above equation, bases are the same. So equating the powers, we get: x = −2

Part (e)
1. Let $\log_9 \frac{1}{531441}$ = x
2. Then we can write: $\frac{1}{531441}$ = 9^x.
⇒ (531441)⁻¹ = 9^x.
3. We have: 531441 = 9⁶.
⇒ (531441)⁻¹ = (9⁶)⁻¹ = 9⁻⁶.
4. From (2) and (3), we get: 9^x = 9⁻⁶.
• In the above equation, bases are the same. So equating the powers, we get: x = −6

Part (f)
1. Let $\log_{\frac{3}{2}} \frac{27}{8}$ = x
2. Then we can write: $\frac{27}{8} = \left(\frac{3}{2} \right)^x$.
3. We have: $\frac{27}{8} = \left(\frac{3}{2} \right)^3$.
4. From (2) and (3), we get: $\left(\frac{3}{2} \right)^x = \left(\frac{3}{2} \right)^3 $.
• In the above equation, bases are the same. So equating the powers, we get: x = 3

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Now we will see how logarithm can be used as a function. It can be written in 4 steps:
1. We know that, if y = b^x, then: log_b y = x.
2. Consider the expression log_b y = x.
• Here, the input y is being processed to obtain y.
• In other words, y is being subjected to a process. The process is nothing but "finding the logarithm of y". (The solved examples that we saw just above, show us how to find the logarithm of a given number). The resulting logarithm is the output.
3. But for functions,
    ♦ Input values are denoted as x.
      ✰ They are plotted along the x-axis.
    ♦ Output values are denoted as y.
      ✰ They are plotted along the y-axis
• So in the expression log_b y = x, we need to interchange x and y. We get: log_b x = y.
4. So we get a function in which, input x is processed to give output y.
• The process is nothing but "finding the logarithm of x". The resulting logarithm is the output y.
• We can write: y = f(x) = log_b x.
This is called logarithmic function.

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Let us write the important features about logarithmic functions. It can be written in 7 steps:
1. The general form of the logarithmic function is: f(x) = log_b x.
   ♦ b should be greater than zero.
   ♦ b should not be equal to 1.

2. Let us see why b should be greater than zero. It can be written in (iii) steps.
(i) Suppose that, b = −2.
Then the function will be y = f(x) = log₍₋₂₎ x
(ii) While plotting the graph, when the input is 2, we need to find log₍₋₂₎ 2
• Let us try:
   ♦ Assume log₍₋₂₎ 2 = y
   ♦ Then we can write: 2 = (−2)^y.
   ♦ ⇒ −2 = 2^(1/y).
   ♦ This is impossible because, no power of 2 will give a negative value.
(iii) To avoid such situations, we avoid −ve numbers altogether.

3. Let us see why b should not be equal to one. It can be written in (iii) steps.
(i) Suppose that, b = 1.
Then the function will be y = f(x) = log₁ x
(ii) While plotting the graph, when the input is 2, we need to find log₁ 2
• Let us try:
   ♦ Assume log₁ 2 = y
   ♦ Then we can write: 2 = (1)^y.
   ♦ This is impossible because, all powers of 1 give 1.
(iii) To avoid such a situation, we avoid 1.

4. Fig.21.18 below shows the graphs of some simple logarithmic functions.

Fig.21.18

• Red, and yellow belong to the category: b > 1
• Green and magenta belong to the category: 0 < b < 1
• White shows b = e. It belong to the category: b > 1

5. From the graphs, we see that,
• When b >1:
   ♦ As x increases towards ∞, f(x) also approaches ∞.
         ✰ Red, yellow and white are rising up.
   ♦ As x decreases towards zero, f(x) approaches −∞.
         ✰ Red, yellow and white are falling down.
• When 0 < b < 1:
    ♦ As x increases towards ∞, f(x) approaches −∞.
          ✰ Green and magenta are falling down.
    ♦ As x decreases towards zero, f(x) approaches ∞.
          ✰ Green and magenta are rising up.


6. From the graphs, we see that:
• Input x cannot be a −ve number.
The reason can be demonstrated in (iii) steps:
(i) Suppose that, x = −1000 and b = 10
(Recall that, b cannot be −ve)
So we want to find y, where y = log₁₀ (−1000)
(ii) Let us try:
   ♦ We can write: −1000 = (10)^y.
   ♦ This is impossible because, no power of 10 will give a negative value.
(iii) So for logarithmic functions, input can never be −ve.
We will see the actual proof in higher classes.

• Since no input can be −ve, we say that:
Domain of the logarithmic function is R⁺.

7. From the graphs, we get the following information also:
• The output of a log function can never be zero.
   ♦ Note that, none of the graphs touch the y-axis.
• The range of a log function is (−∞,∞).
• For a log function, the point (1,0) will be always available.
   ♦ This is because, any number raised to the power zero, is 1. Which implies: Whatever be the base, logarithm of 1 is zero.

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Now we will discuss the fact that, log function is the inverse of the exponential function. It can be written in 4 steps:
1. Fig.21.19 shows the graphs of three functions:
(i) y = f(x) = e^x. (red color)
(ii) y = f(x) = x. (magenta color)
(iii) y = f(x) = log_e x. (yellow color)

Fig.21.19

2. Mark any point P on the magenta line.
• Through P, draw a white dashed line perpendicular to the magenta line.
• The white dashed line,
   ♦ intersects the red curve at Q.
   ♦ intersects the yellow curve at R.
• The distances PQ and PR are equal.

3. PQ and PR are equal because,
   ♦ red and yellow curves are mirror images of each other.
   ♦ magenta line is the mirror line.

4. y = e^x is the inverse of y = log_e x and vice versa.
If we combine them as a composite function, we will get an identity function. This can be demonstrated in (iii) steps:
(i) Let f(x) = e^x and g(x) = log_e x.
• Then f(g(x)) = f(log_e x) = $\rm{e^{\log_e x}}$
• Let $\rm{e^{\log_e x}}$ = u
⇒ log_e u = log_e x
⇒ u = x
⇒ f(g(x)) = u = x
(ii) Similarly, g(f(x)) = g(e^x) = log_e (e^x).
• Let log_e (e^x) = v
⇒ e^x = e^v.
⇒ v = x
⇒ g(f(x)) = v = x
(iii) From (i) and (ii), we get: f(g(x)) = g(f(x)) = x
• That means, f(x) is the inverse of g(x) and vice versa.
• In other words, y = e^x is the inverse of y = log_e x and vice versa.
• This true for all acceptable values of base b. We can write:
   ♦ y = 10^x is the inverse of y = log₁₀ x and vice versa.
   ♦ y = 2^x is the inverse of y = log₂ x and vice versa.
   ♦ y = e^x is the inverse of y = log_e x and vice versa.
   ♦ etc.,

Based on this information, let us see a solved example:

Solved example 21.42
Is it true that $\rm{x = e^{\log x}}$ for all real x?
Solution:
1. Consider $\rm{x = e^{\log x}}$.
• It is a composite of two functions:
    ♦ $\rm{y = e^x}$
    ♦ $\rm{y = \log_e x}$
(Recall that, in this chapter, we write $\rm{\log_e x}$ simply as $\rm{\log x}$)
2. Each of the two functions is inverse of the other.
• So whatever is the input, the output will be same as input.
3. Now suppose that, the input x is a −ve number.
• Then the "$\rm{\log x}$" portion of the composite function will not be able to process the input x. This is because, logarithm of −ve numbers does not exist.
4. Therefore, the equation $\rm{x = e^{\log x}}$ is true only when input x is +ve.

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In a logarithm function, if e is taken as the base, then we denote it as ln. This can be explained in 4 steps:
1. Recall that, in the case of the exponential function y = b^x, if the base is 10, we call it common exponential function.
• In a similar way, in the case of the logarithmic function y = log_b x, if the base is 10, we call it common logarithmic function.
2. Recall that, in the case of the exponential function y = b^x, if the base is e, we call it natural exponential function.
• In a similar way, in the case of the logarithmic function y = log_b x, if the base is e, we call it natural logarithmic function.
3. So the natural logarithmic function is: y = log_e x
• It can be written in a short form as: y = ln x.
4. In this chapter, if the base is not specified, it means natural logarithm.
• So, in this chapter, if we see y = log 5, then it means, y = the natural logarithm of 5.

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We have completed a basic discussion on logarithmic functions. In the next section, we will see properties of logarithmic functions.

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21.11 - Exponential Functions

In the previous section, we completed a discussion on derivatives of inverse trigonometric functions. In this section, we will see exponential functions.

First we will see some basics about polynomial functions. It can be written in 4 steps:
1.Consider the following five polynomial functions:
(i) y = f₁(x) = x¹
(ii) y = f₂(x) = x²
(iii) y = f₃(x) = x³
(iv) y = f₄(x) = x⁴
(v) y = f₅(x) = x⁵
• They have the general form y = f_n(x) = xⁿ

2. Fig.21.15 below shows the graphs of these functions.

Fig.21.15

• We see that, as n increases, the curves become steeper.
   ♦ "Steeper" means, the graph leans more towards the y-axis.
   ♦ "Flatter" means, the graph leans more towards the x-axis.

3. For a steeper curve, growth will be faster. That is., even for a small increase in x, there will be a large increase in y.
• This can be demonstrated by comparing two graphs. The comparison can be written in (v) steps:
(i) Let us compare the graphs of
   ♦ y = f₁₀(x) = x¹⁰  and
   ♦ y = f₁₅(x) = x¹⁵
• 15 > 10. So y = f₁₅(x) = x¹⁵ will be steeper than y = f₁₀(x) = x¹⁰.
(ii) Consider y = f₁₀(x) = x¹⁰.
• Beginning from x = 2, let us move a distance of 3 units along the x axis, towards the right.
• So we begin from x = 2 and end at x = 5.
   ♦ At x = 2, y = 2¹⁰. 
   ♦ At x = 5, y = 5¹⁰. 
• The corresponding increase in y is:
(5¹⁰ − 2¹⁰) = (9765625 − 1024) = 9764601
(iii) Consider y = f₁₅(x) = x¹⁵.
• Let us move the same distance, from the same point, in the same direction.
• So we begin from x = 2 and end at x = 5.
   ♦ At x = 2, y = 2¹⁵. 
   ♦ At x = 5, y = 5¹⁵. 
• The corresponding increase in y is:
(5¹⁵ − 2¹⁵) = (30517578125 − 32768) = 30517545357.
(iv) Comparing 9764601 and 30517545357, we can write:
• For the same increase in x,
   ♦ y = f₁₅(x) = x¹⁵
   ♦ gives a greater increase in y than
   ♦ y = f₁₀(x) = x¹⁰.
(v) So we can conclude that:
• For the same increase in x,
   ♦ the steeper curve
   ♦ gives a greater increase in y than
   ♦ the flatter curve.

4. Based on the above comparison, we can write:
Rate of growth of a polynomial function is dependent on the degree of that polynomial function. Higher the degree, higher is the rate of growth. We will see the actual proof in higher classes.

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• Consider a polynomial function with a very large degree. Even such a large degree will not be sufficient to describe some situations in science and engineering. Loudness of sound, radioactive decay, wave lengths in electromagnetic spectrum, are examples.
• In such situations, we use exponential functions.
• Exponential functions have the general form: y = f(x) = b^x.
   ♦ where b > 0 and b ≠ 1.
   ♦ As a sample, y = f(x) = 10^x is plotted in the fig.21.15 above.
• Rate of growth of an exponential function, will be greater than that of any polynomial function. This is true when x is very large.
• This fact can be demonstrated by a simple comparison. It can be written in (v) steps:
(i) Let us compare the graphs of
   ♦ y = f₁₀₀(x) = x¹⁰⁰  and
   ♦ y = f(x) = 10^x
(ii) Consider y = f₁₀₀(x) = x¹⁰⁰.
• Let us move from x = 10³ to x = 10⁵
   ♦ At x = 10³, y = (10³)¹⁰⁰ = 10³⁰⁰. 
   ♦ At x = 10⁵, y = (10⁵)¹⁰⁰ = 10⁵⁰⁰. 
• The corresponding increase in y is:
(10⁵⁰⁰ − 10³⁰⁰)
(iii) Consider y = f(x) = 10^x.
• Let us move the same distance, from the same point, in the same direction.
   ♦ At x = 10³, y = (10)¹⁰³ = 10¹⁰⁰⁰. 
   ♦ At x = 10⁵, y = (10)¹⁰⁵ = 10¹⁰⁰⁰⁰⁰. 
• The corresponding increase in y is:
(10¹⁰⁰⁰⁰⁰ − 10¹⁰⁰⁰).
(iv) Comparing (10⁵⁰⁰ − 10³⁰⁰) and (10¹⁰⁰⁰⁰⁰ − 10¹⁰⁰⁰), we can write:
• For the same increase in x,
   ♦ y = f(x) = 10^x
   ♦ gives a greater increase in y than
   ♦ y = f₁₀₀(x) = x¹⁰⁰.
(v) So we can conclude that:
• For the same increase in x,
   ♦ the exponential function
   ♦ gives a greater increase in y than
   ♦ the polynomial function.

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Let us write the important features about exponential functions. It can be written in 6 steps:
1. The general form of the exponential function is: f(x) = b^x
   ♦ b should be greater than zero.
   ♦ b should not be equal to 1.

2. Let us see why b should be greater than zero. It can be written in (iii) steps.
(i) Suppose that, b = −9.
Then the function will be y = f₍₋₉₎ = (−9)^x
(ii) While plotting the graph, when the input is 1/2, we get:
$f_{(-9)} = (-9)^{\frac{1}{2}} = \sqrt{-9} = 3i$
• This is not defined because, $3i$ is not a real number.
(iii) To avoid such situations, we avoid −ve numbers altogether.

3. Let us see why b should not be equal to one. It can be written in (iii) steps.
(i) Suppose that, b = 1.
Then the function will be y = f₁ = 1^x
(ii) While plotting the graph, whatever is the input, we will get y = f₁(x) = 1
• This is a constant function. A constant function is not useful for describing variation of quantities like loudness of sound, wavelengths in the electromagnetic spectrum etc,.
(iii) To avoid such a situation, we avoid 1.

4. Fig.21.16 below shows the graphs of some simple exponential functions.

Fig.21.16

• Red, yellow and magenta belong to the category: b > 1
• Green, pink and orange belong to the category: 0 < b < 1

 5. From the graphs, we see that,
• When b >1:
   ♦ As x increases towards ∞, f(x) also approaches ∞.
         ✰ Red, yellow and magenta are rising up. 
   ♦ As x decreases towards −∞, f(x) approaches zero. 
         ✰ Red, yellow and magenta are approaching x-axis. 
• When 0 < b < 1:
   ♦ As x increases towards ∞, f(x) approaches zero.
         ✰ Green, pink and orange are approaching x-axis.
♦ As x decreases towards −∞, f(x) approaches ∞.
         ✰ Green, pink and orange are rising up.

6. From the graphs, we get the following information also:
• Any real number can be used as an input for exponential function. That means, the domain is (−∞,∞)
• The output of an exponential function can never be zero. It can only approach zero.
• The output of an exponential function will be always +ve.
• The range of an exponential function is (0,∞).
• For an exponential function, the point (0,1) will be always available. This is because, any number raised to the power zero, is 1.

---

Now we will see common exponential function. It can be explained in 2 steps:
1. We know that, the general form of exponential function is: f(x) = b^x.
• "b" is called base of the exponential function.
2. If b = 10, then we get: f(x) = 10^x.
• This function is known as common exponential function.

---

Finally we will see natural exponential function. It can be explained in 5 steps:
1. We know that, the general form of exponential function is: f(x) = b^x.
• "b" is called base of the exponential function.
2. If b = e, then we get: f(x) = e^x.
• This function is known as natural exponential function.
3. Recall that, e is the sum of the infinite series:
$1 + \frac{1}{1!}+ \frac{1}{2!} + ~.~.~.$ (Details here)
• The value of e is 2.71828182845905…
4. We see that, e lies between 2 and 3. So the graph of the natural exponential function will lie between the graphs of y = f₂ = 2^x and y = f₃ = 3^x .
• It is shown in the fig.21.17 below:

Fig.21.17

5. Many problems in science, engineering and economics give simple solutions if e is used as the base. Instead of e, if 10 is used, those solutions will become large and complicated.

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We have completed a discussion on exponential functions. In the next section, we will see logarithmic functions.

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21.10 - Derivatives of Inverse Trigonometric Functions

In the previous section, we completed a discussion on derivatives of implicit functions. In this section, we will see derivatives of inverse trigonometric functions.

• We have seen inverse trigonometric functions in chapter 18. Now we want to find the derivatives of those functions.
• We know that, only continuous functions are differentiable. Inverse trigonometric functions are continuous functions. We will see the proof in higher classes.
• At present we will see the method for differentiating those functions.

Let us see an example. It can be written in 8 steps:
1. Given that, f(x) = sin⁻¹x. We want to find f'(x).
2. Let y = sin⁻¹x. Then x = sin y
3. Differentiating both sides with respect to x, we get:
$\frac{d}{dx} (x) ~=~ \frac{d}{dx} (\sin y)$
⇒ $1~=~ \cos y \frac{dy}{dx}$
⇒ $\frac{dy}{dx}~=~\frac{1}{\cos y}~=~\frac{1}{\cos(\sin^{-1} x)}$
4. In the above result, the denominator cos y should not be equal to zero.
• That means, y should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• That means, sin⁻¹x should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• That means, x should not be equal to −1 or 1. 
5. In chapter 18, we saw that, the acceptable domain for the sin⁻¹ function is [−1,1] (details here).
• But from the above step (4), we see that, derivative of the sin⁻¹ function is not defined at −1 and 1.
• So we can write two important points:
(i) For the sin⁻¹ function, the input can be taken from the interval [−1,1].
(ii) Derivative of the sin⁻¹ function is available only in the interval (−1,1).
6. We can eliminate the trigonometric ratios from the result.
• We have:
$\cos^2 y ~=~ 1 - \sin^2 y ~=~ 1 - [\sin(\sin^{-1} x)]^2 ~=~1 - x^2 $
So $\cos y ~=~ \pm \sqrt{1 - x^2}$
7. We have to determine whether cos y is $\sqrt{1 - x^2}$ or $- \sqrt{1 - x^2}$.
• The input for cos y is "y", which is "sin⁻¹x".
• We saw that, sin⁻¹x should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• That means, input for cos y should not be equal to $-\frac{\pi}{2}$ or $\frac{\pi}{2}$.
• Between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, cosine is +ve.
• So cos y is +ve.
• Thus $\cos y ~=~ \sqrt{1 - x^2}$
8. So we can write:
If y = sin⁻¹ x, then $\frac{dy}{dx}~=~\frac{1}{\cos y}~=~\frac{1}{\cos(\sin^{-1} x)}~=~ \frac{1}{\sqrt{1 - x^2}}$

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Now we will see the derivative of inverse cosine function. It can be written in 8 steps:

1. Given that, f(x) = cos⁻¹x. We want to find f'(x).
2. Let y = cos⁻¹x. Then x = cos y
3. Differentiating both sides with respect to x, we get:
$\frac{d}{dx} (x) ~=~ \frac{d}{dx} (\cos y)$
⇒ $1~=~ -\sin y \frac{dy}{dx}$
⇒ $\frac{dy}{dx}~=~\frac{-1}{\sin y}~=~\frac{-1}{\sin(\cos^{-1} x)}$
4. In the above result, the denominator sin y should not be equal to zero.
• That means, y should not be equal to 0 or $\pi$.
• That means, cos⁻¹x should not be equal to 0 or $\pi$.
• That means, x should not be equal to 1 or −1. 
5. In chapter 18, we saw that, the acceptable domain for the cos⁻¹ function is [−1,1] (details here).
• But from the above step (4), we see that, derivative of the cos⁻¹ function is not defined at −1 and 1.
• So we can write two important points:
(i) For the cos⁻¹ function, the input can be taken from the interval [−1,1].
(ii) Derivative of the cos⁻¹ function is available only in the interval (−1,1).
6. We can eliminate the trigonometric ratios from the result.
• We have:
$\sin^2 y ~=~ 1 - \cos^2 y ~=~ 1 - [\cos(\cos^{-1} x)]^2 ~=~1 - x^2 $
So $\sin y ~=~ \pm \sqrt{1 - x^2}$
7. We have to determine whether sin y is $\sqrt{1 - x^2}$ or $- \sqrt{1 - x^2}$.
• The input for sin y is "y", which is "cos⁻¹x".
• We saw that, cos⁻¹x should not be equal to 0 or π.
• That means, input for sin y should not be equal to 0 or π.
• Between 0 and π, sine is +ve.
• So sin y is +ve.
• Thus $\sin y ~=~ \sqrt{1 - x^2}$
8. So we can write:
If y = cos⁻¹ x, then $\frac{dy}{dx}~=~\frac{-1}{\sin y}~=~\frac{-1}{\sin(\cos^{-1} x)}~=~\frac{-1}{\sqrt{1 - x^2}}$

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Now we will see the derivative of inverse tangent function. It can be written in 4 steps:

1. Given that, f(x) = tan⁻¹x. We want to find f'(x).
2. Let y = tan⁻¹x. Then x = tan y
3. Differentiating both sides with respect to x, we get:
$\frac{d}{dx} (x) ~=~ \frac{d}{dx} (\tan y)$
⇒ $1~=~ \sec^2 y \frac{dy}{dx}$
⇒ $\frac{dy}{dx}~=~\frac{1}{\sec^2 y}~=~\frac{1}{1 + \tan^2 y}~=~\frac{1}{1 + x^2 }$
4. In the above result, the denominator 1 + x² can never become zero.
• That means, x can be any real number.

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• We have seen the derivatives of:
    ♦ inverse sine function
    ♦ inverse cosine function
    ♦ inverse tangent function

• There are three more inverse trigonometric functions:
    ♦ inverse cosecant function
    ♦ inverse secant function
    ♦ inverse cotangent function
        ✰ Derivatives of these three functions can be obtained by writing them in terms of sine, cosine or tangent. For that, we can use the first property of inverse trigonometric functions. Details here.

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Let us see some solved examples.

Solved example 21.37
Find $\frac{dy}{dx}$ if y = 4 cos⁻¹ x − 10 tan⁻¹ x.
Solution:

Solved example 21.38
Find $\frac{dy}{dx}~~ \text{if}~~ y \,=\, \sqrt{x} \sin^{-1} x$.
Solution:

 

Solved example 21.39
Find $\frac{dy}{dx}~~ \text{if}~~ y \,=\, \frac{1}{\sin^{-1} x}$
Solution:

Solved example 21.40
Find $\frac{dy}{dx}~~ \text{if}~~ y \,=\, x \tan^{-1} \sqrt{x}$
Solution:

 

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Link to a few more solved examples is given below:

Exercise 21.3

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We have completed a discussion on the derivatives of inverse trigonometric functions. In the next section, we will see Exponential functions.

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21.9 - Derivatives of Implicit Functions - Easy method

In the previous section, we saw the method to find the derivatives of implicit functions. In this section, we will see some solved examples. In those solved examples, we will see an easy method also.

Solved example 21.32
Find $\frac{dy}{dx}$ if x + sin (xy) − y = 0.
Solution:
1. Consider the function x + sin (xy) − y = 0.
• We want the derivative of this function with respect to x.
   ♦ That is., we want $\frac{dy}{dx}$.
2. Consider the "y" in the given equation. It is a function of "x" because, value of y depends on the value of x. So we can write "f(x)" in the place of "y". We get:
x + sin (x f(x)) − f(x) = 0
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (x) \,+\, \frac{d}{dx} \sin(x f(x)) \, - \, \frac{d}{dx} f(x) \,=\, \frac{d}{dx} (0)$
⇒ $1 \,+\, \frac{d}{dx} \sin(x f(x)) \, - \, f'(x) \,=\, 0$
4. The second term in the L.H.S needs special attention. It is the derivative of a composite function. It can be determined in 4 steps:
(i) g(x) = sin (x f(x)) = (v∘u)(x) = v(u(x))
Where u(x) = x f(x) and v(u(x)) = sin (x f(x)).
(ii) v'(u(x)) = cos (x f(x))
(iii) u'(x) = 1.f(x) + x f'(x)
(iv) So g'(x) = v'(u(x)).u'(x)
= f(x).cos (x f(x)) + x f'(x) cos(x f(x))
5. So the result in (3) becomes:
$1 \,+\, f(x).\cos (x f(x)) + x. f'(x). \cos(x f(x)) \, - \, f'(x) \,=\, 0$
⇒ $1 \,+\, f(x).\cos (x f(x)) ~+~  f'(x) \left[x. \cos(x. f(x)) \, - \, 1 \right] \,=\, 0$
⇒ $1 \,+\, f(x).\cos (x f(x)) ~-~  f'(x) \left[1~-~x. \cos(x. f(x))  \right] \,=\, 0$
⇒ $f'(x) = \frac{1 \,+\, f(x).\cos (x. f(x))}{1~-~x. \cos(x. f(x))}$
6. We wrote "f(x)" in the place of "y".
• So f'(x) is $\frac{dy}{dx}$
• Therefore, $\frac{dy}{dx} = \frac{1 \,+\, y \cos (xy)}{1~-~x \cos(xy)}$

---

Once we become familiar with the basics of this method,
    ♦ we need not write "f(x)" in the place of "y". 
        ✰ we can keep "y" as such.
    ♦ we need not write "f'(x)" to show differentiation of y.
        ✰ we can write $\frac{dy}{dx}$ instead.
    ♦ Also in simple cases, we may be able to apply chain rule in just one step.

So let us redo the above solved example.

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} (x)\,+\,\frac{d}{dx} (\sin(xy))\,-\,\frac{d}{dx} (y)}    & {~=~}    &{\frac{d}{dx} (0)}    \\
{~\color{magenta}    2    }    &{\implies}    &{1\,+\,\frac{d}{dx} (\sin(xy))\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{1\,+\,\cos(xy) \left[\frac{d}{dx} (x) .y + x.\frac{dy}{dx} \right]\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{1\,+\,\cos(xy) \left[y + x.\frac{dy}{dx} \right]\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{1\,+\,y \cos(xy) + x.\frac{dy}{dx} \cos(xy)\,-\,\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{\implies}    &{1\,+\,y \cos(xy) + \frac{dy}{dx} \left[x \cos(xy)\,-\,1 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{\implies}    &{1\,+\,y \cos(xy)}    & {~=~}    &{\frac{dy}{dx} \left[1 - x \cos(xy) \right]}    \\
{~\color{magenta}    8    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1\,+\,y \cos(xy)}{1 - x \cos(xy)}}    \\
\end{array}$

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Solved example 21.33
Find $\frac{dy}{dx}$ if x³y⁵ + 3x = 8y³ + 1.
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} (x^3 y^5)\,+\,\frac{d}{dx} (3x)}    & {~=~}    &{\frac{d}{dx} (8 y^3)\,+\,\frac{d}{dx} (1)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{d}{dx} (x^3 y^5)\,+\,3}    & {~=~}    &{\frac{d}{dx} (8 y^3)\,+\,0}    \\
{~\color{magenta}    3    }    &{\implies}    &{3 x^2 . y^5 \,+\,x^3 . \frac{d}{dx} (y^5) \,+\,3}    & {~=~}    &{8 \frac{d}{dx} (y^3)}    \\
{~\color{magenta}    4    }    &{\implies}    &{3 x^2 . y^5 \,+\,x^3 . \left[5 y^4 .  \frac{dy}{dx} \right] \,+\,3}    & {~=~}    &{8. 3 y^2 . \frac{dy}{dx}}    \\
{~\color{magenta}    5    }    &{\implies}    &{3 x^2 y^5 \,+\, 5 x^3 y^4 \frac{dy}{dx} \,+\, 3}    & {~=~}    &{24 y^2 \frac{dy}{dx}}    \\
{~\color{magenta}    6    }    &{\implies}    &{3 x^2 y^5  \,+\, 3}    & {~=~}    &{24 y^2 \frac{dy}{dx} \,-\, 5 x^3 y^4 \frac{dy}{dx}}    \\
{~\color{magenta}    7    }    &{\implies}    &{3 x^2 y^5  \,+\, 3}    & {~=~}    &{\left[24 y^2  \,-\, 5 x^3 y^4 \right]\frac{dy}{dx}}    \\
{~\color{magenta}    8    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{3 x^2 y^5  \,+\, 3}{24 y^2  \,-\, 5 x^3 y^4}}    \\
\end{array}$

◼ Remarks:
• 3 (magenta color). For differentiating x³y⁵, we use product rule. For differentiating y⁵, we treat it as a composite function.

• In the R.H.S, y³ is also treated as a composite function.

Solved example 21.34
Find $\frac{dy}{dx}$ if x² + xy + y² = 100.
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} (x^2)\,+\,\frac{d}{dx} (xy) \,+\,\frac{d}{dx} (y^2)}    & {~=~}    &{\frac{d}{dx} (100)}    \\
{~\color{magenta}    2    }    &{\implies}    &{2x\,+\,\frac{d}{dx} (xy) \,+\,\frac{d}{dx} (y^2)}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{2x\,+\,\left[1. {y} + x \frac{dy}{dx} \right] \,+\,\left[2y \frac{dy}{dx} \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{2x + y + (x+2y)\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{2x + y}    & {~=~}    &{- (x+2y)\frac{dy}{dx}}    \\
{~\color{magenta}    6    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{-(2x + y)}{(x+2y)}}    \\
\end{array}$

Solved example 21.35
Find $\frac{dy}{dx}$ if x³ + x²y + xy² + y³ = 81.
Solution:

Solved example 21.36
Find $\frac{dy}{dx}$ if sin²y + cos xy = π.
Solution:

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We have completed a discussion on the derivatives of implicit functions. In the next section, we will see derivatives of inverse trigonometric functions.

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21.8 - Derivatives of Implicit Functions

In the previous section, we completed a discussion on chain rule. We saw some solved examples also. In this section, we will see derivatives of implicit functions.

First we will see some basic details about explicit functions. It can be written in 7 steps:
1. Consider the equation: x−y−π = 0
2. This equation has two variables x and y. The equation can be rearranged and written as:
y = x − π
After the rearrangement, the equation has become a function. We see that, y is a function of x.
3. We can write either of the two ways:
    ♦ y = x − π
    ♦ f(x) = x − π
Now it is easy to find the derivative $\frac{dy}{dx}$ or f'(x).
4. So there are functions which can be easily rearranged so that three conditions are satisfied:
(i) There is a single "y" variable on the L.H.S.
(ii) All "x terms" are on the R.H.S.
(iii) There is no "y" variable or "y term" on the R.H.S.
• Such functions are called explicit functions.
5. The equation xy = 1 is another example.
• We can rearrange this equation and write: $y=\frac{1}{x}$
• All three conditions that we wrote in (4) are satisfied.
• So xy = 1 is an explicit function.
6.In the case of explicit functions, We say that:
y is given as an explicit function of x
7. The dictionary meaning of 'explicit' is:
Clear and exact. There will not be any need for doubts.
• Indeed, in the case of explicit functions, we can clearly see how 'y' is dependent on 'x'.

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Now we will see some basic details about implicit functions. It can be written in 5 steps:
1. Consider the equation: x+sin xy −y = 0
2. This equation has two variables x and y. The equation cannot be easily rearranged so as to satisfy the three conditions that we wrote in the case of explicit functions.
• Such functions are called implicit functions.
3. In the case of implicit functions also, "y" is dependent on "x". But the manner of this dependency is not clear and exact.
• We say that:
y is given as an implicit function of x.
That is., dependency of 'y' on 'x' is implied.
4. The dictionary meaning of 'implicit' is:
A fact is implied, but not communicated directly.
5. The equation x³y⁵ + 3x = 8y³ + 1
is another example.
• We cannot rearrange it easily.

---

Let us see how to find the derivative in the case of implicit functions. It can be explained in steps:

So far in this chapter, we have been finding the derivatives of explicit functions.
Consider the explicit function f(x) = 2x+3.
• This can be written as y = 2x+3 also.
• Using 'y' instead of 'f(x)' is useful when we plot the graph of the function.
   ♦ We plot the input values along the x-axis.
   ♦ The output values are 'f(x) values'. They are plotted along the y-axis.
   ♦ So using 'y' instead of 'f(x)' will not make any difference.

• We will now see a new method for finding the derivative. It can be explained using an example.
Let us first use an explicit function as example. The method can be written in 5 steps:
1. Consider the function x−y = π.
• We want the derivative of this function with respect to x.
   ♦ That is., we want $\frac{dy}{dx}$.
2. Consider the "y" in the given equation. It is a function of "x" because, value of y depends on the value of x. So we can write "f(x)" in the place of "y". We get:
x−f(x) = π
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (x) \,-\, \frac{d}{dx} f(x) \,=\, \frac{d}{dx} (\pi)$
⇒ 1 − f'(x) = 0
⇒ f'(x) = 1
4. We wrote "f(x)" in the place of "y".
• So f'(x) is $\frac{dy}{dx}$
• Therefore, $\frac{dy}{dx}$ = 1 
5. We will get the same result even if we begin with y = x−π.

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Let us see another example to demonstrate the new method. This time also, we will use an explicit function.

1. Consider the function xy = 1.
• We want the derivative of this function with respect to x.
   ♦ That is., we want $\frac{dy}{dx}$.
2. Consider the "y" in the given equation. It is a function of "x" because, value of y depends on the value of x. So we can write "f(x)" in the place of "y". We get:
x f(x) = 1
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (x) . f(x) \,+\, x \frac{d}{dx} f(x) \,=\, \frac{d}{dx} (1)$
⇒ 1.f(x) + x f'(x) = 0
⇒ f'(x) = $-\frac{f(x)}{x}$
4. We wrote "f(x)" in the place of "y".
• So f'(x) is $\frac{dy}{dx}$
• Therefore, $\frac{dy}{dx} = -\frac{f(x)}{x} = - \frac{y}{x} = - \frac{1/x}{x} = -\frac{1}{x^2}$
5. We will get the same result even if we begin with $y=\frac{1}{x}$.

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Now we will apply the new method to an implicit function.

1. Consider the function y + sin y = cos x.
• We want the derivative of this function with respect to x.
   ♦ That is., we want $\frac{dy}{dx}$.
2. Consider the "y" in the given equation. It is a function of "x" because, value of y depends on the value of x. So we can write "f(x)" in the place of "y". We get:
f(x) + sin (f(x)) = cos x
(We need not worry about how f(x) is defined. This fact will become clear when we solve a few problems)
3. Differentiating both sides, we get:
$\frac{d}{dx} (f(x)) \,+\, \frac{d}{dx} \sin(f(x)) \,=\, \frac{d}{dx} (\cos x)$
⇒ $f'(x) \,+\, \frac{d}{dx} \sin(f(x)) \,=\, - \sin x$
4. The second term in the L.H.S needs special attention. It is the derivative of a composite function. It can be determined in 4 steps:
(i) g(x) = sin (f(x)) = (v∘u)(x) = v(u(x))
Where u(x) = f(x) and v(u(x)) = sin (f(x)).
(ii) v'(u(x)) = cos (f(x))
(iii) u'(x) = f'(x)
(iv) So g'(x) = v'(u(x)).u'(x) = cos (f(x)).f'(x)
5. So the result in (3) becomes:
$f'(x) \,+\,  \cos(f(x)).f'(x) \,=\, - \sin x$
⇒ $f'(x) \left[1 \,+\,  \cos(f(x)) \right] \,=\, - \sin x$
⇒ $f'(x) \,=\, - \frac{\sin x}{1 \,+\,  \cos(f(x))}$
6. We wrote "f(x)" in the place of "y".
• So f'(x) is $\frac{dy}{dx}$
• Therefore, $\frac{dy}{dx} = - \frac{\sin x}{1 \,+\,  \cos(f(x))} = - \frac{\sin x}{1 \,+\,  \cos y}$
• Here, the denominator should not be zero. That means, cos y should not be −1.
• So we can write: y ≠ (2n+1)π

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

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21.7 - Chain Rule - Easy Method

In the previous section, we saw the basics about chain rule. We saw some solved examples also. In this section, we will see an alternate method for applying this rule.

The alternate method can be explained in 6 steps:
1. Recall the first example that we saw in the previous section.
• If f(x) = (3x+4)³, then
$\frac{df}{dx} \,=\, 9 (3x+4)^2$
2. The above result can be split up as:
$\frac{df}{dx} \,=\, 9 (3x+4)^2\,=\,\left[3 (3x+4)^2 \right].3$
So the result has two parts:
(i) $\left[3 (3x+4)^2 \right]$
(ii) 3
3. The function in our example is: f(x) = (3x+4)³.
We can write it as the composite of two functions: f(x) = v(u(x))
• Where
     (i) u(x) = 3x+4
     (ii) v(u(x)) = [u(x)]³.
• Then f(x) = v(u(x)) = v(3x+4) =  (3x+4)³.
4. Consider 3(ii) above. From that equation, we get:
$v'(u(x))\,=\,3[u(x)]^2\,=\,3 (3x+4)^2$
• So part (i) of the result that we wrote in step 2 is v'(u(x)).
5. Consider 3(i) above. From that equation, we get:
$u'(x)\,=\,3$
• So part (ii) of the result that we wrote in step 2 is u'(x).
6. Based on the above steps, we can write the procedure in brief:
(i) We are given f(x) = v(u(x)). We want f'(x).
(ii) First consider u(x) as a whole, and differentiate it. That means, we are finding v'(u(x))
(iii) Then differentiate u(x). That means, we are finding u'(x)
(iv) Finally, multiply the results in (ii) and (iii).
6. If it is a composite of 3 functions, then also, we can write the procedure in brief:
(i) We are given f(x) =w(v(u(x))). We want f'(x).
(ii) First consider v(u(x)) as a whole, and differentiate it. That means, we are finding w'(v(u(x)))
(iii) Next consider u(x) as a whole, and differentiate it. That means, we are finding v'(u(x))
(iv) Next differentiate u(x). That means, we are finding u'(x)
(v) Finally, multiply the results in (ii), (iii) and (iv).

---

Let us revisit the solved examples that we saw in the previous section.

Solved example 21.21
Find the derivative of the function given by f(x) = sin (x²)
Solution (alternate method):
1. f(x) = sin (x²) = (v∘u)(x) = v(u(x))
Where u(x) = x² and v(u(x)) = sin (u(x)).
2. v'(u(x)) = cos (x²)
3. u'(x) = 2x
4. So f'(x) = v'(u(x)).u'(x) = 2x cos (x²)

Solved example 21.22
Find the derivative of the function given by f(x) = tan (2x+3)
Solution (alternate method):
1. f(x) = tan (2x+3) = (v∘u)(x) = v(u(x))
Where u(x) = 2x+3 and v(u(x)) = tan (u(x)).
2. v'(u(x)) = sec²(2x+3)
3. u'(x) = 2
4. So f'(x) = v'(u(x)).u'(x) = 2 sec²(2x+3)

Solved example 21.23
Find the derivative of the function given by f(x) = sin (cos (x²))
Solution (alternate method):
1. f(x) = sin (cos (x²)) = (w∘v∘u)(x) = (w∘v)(u(x))
Where u(x) = x²,  v(u(x)) = cos (u(x)) and w(v(u(x))) = sin (v(u(x))).
2. w'(v(u(x))) = cos (cos (x²))
3. v'(u(x)) = −sin (x²)
4. u'(x) = 2x
5. So f'(x) = w'(v(u(x))).v'(u(x)).u'(x)
= −2x sin (x²) cos (cos (x²))

Solved example 21.24
Find the derivative of the function given by
$f(x) = \sqrt{5x - 8}$
Solution (alternate method):
1. $f(x) = \sqrt{5x - 8}$ = (v∘u)(x) = v(u(x))
Where u(x) = 5x −8 and v(u(x)) = $\sqrt{u(x)}$.
2. v'(u(x)) = $\frac{1}{2} (5x - 8)^{- \frac{1}{2}}$
3. u'(x) = 5
4. So f'(x) = v'(u(x)).u'(x) = $\frac{5}{2 \sqrt{5x - 8}}$

---

Now we will see some new solved examples:

Solved example 21.25
Find the derivative of the function given by
$f(x) = \sin \left(3x^2 + x \right)$
Solution:
1. $f(x) = \sin \left(3x^2 + x \right)$ = (v∘u)(x) = v(u(x))
Where u(x) = (3x² + x) and v(u(x)) = sin (u(x)).
2. v'(u(x)) = cos (3x² + x)
3. u'(x) = 6x + 1
4. So f'(x) = v'(u(x)).u'(x) = (6x+1) cos (3x² + x)

Solved example 21.26
Find the derivative of the function given by
f(x) = sec (1 − 5x)
Solution:
1. f(x) = sec (1 − 5x) = (v∘u)(x) = v(u(x))
Where u(x) = (1 − 5x) and v(u(x)) = sec (u(x)).
2. v'(u(x)) = sec (1 − 5x) tan (1 − 5x)
3. u'(x) = −5
4. So f'(x) = v'(u(x)).u'(x) = −5sec (1 − 5x) tan (1 − 5x)

Solved example 21.27
Find the derivative of the function given by
$f(x) = \left(2 x^3 \,+\, \cos x \right)^{50}$
Solution:
1. $f(x) = \left(2 x^3 \,+\, \cos x \right)^{50}$ = (v∘u)(x) = v(u(x))
Where $u(x) =\left(2 x^3 \,+\, \cos x \right)$ and $v(u(x)) = (u(x))^{50}$.
2. v'(u(x)) = $50\left(2 x^3 \,+\, \cos x \right)^{49}$
3. To find u'(x):
Part (i): differentiating 2x³:
The answer is 6x².
Part (ii): differentiating cos x:
The answer is −sin x.
• So u'(x) = 6x² − sin x.
4. So $f'(x) = 50 \left(6x^2 - \sin x \right) \left(2 x^3 \,+\, \cos x \right)^{49}$

Solved example 21.28
Find the derivative of the function given by
f(x) = cos² (x)
Solution:
1. f(x) = cos² (x) = (v∘u)(x) = v(u(x))
Where u(x) = cos (x) and v(u(x)) = (u(x))² .
2. v'(u(x)) = 2 u(x) = 2 cos (x)
3. u'(x) = −sin (x)
4. So f'(x) = v'(u(x)).u'(x) = −2 sin (x) cos (x)
(This result can be obtained using product rule also)

Solved example 21.29
Find the derivative of the function given by
f(x) = sin² (x)
Solution:
1. f(x) = sin² (x) = (v∘u)(x) = v(u(x))
Where u(x) = sin (x) and v(u(x)) = (u(x))² .
2. v'(u(x)) = 2 u(x) = 2 sin (x)
3. u'(x) = cos (x)
4. So f'(x) = v'(u(x)).u'(x) = 2 sin (x) cos (x)
(This result can be obtained using product rule also)

Solved example 21.30
Find the derivative of the function given by
f(x) = cos⁴ (x) + cos (x⁴)
Solution:
• The given function is of the form:
f(x) = g(x) + h(x)
• Applying algebra of derivatives, we can write:
f'(x) = g'(x) + h'(x)

Part (i): Finding g'(x)
1. g(x) = cos⁴ (x) = (v∘u)(x) = v(u(x))
Where u(x) = cos² (x) and v(u(x)) = (u(x))² .
2. v'(u(x)) = 2 u(x) = 2 cos² (x)
3. u'(x) = −2 sin (x) cos (x)
4. So f'(x) = v'(u(x)).u'(x) = −4 sin (x) cos³ (x)

Part (ii): Finding h'(x)
1. h(x) = cos (x⁴) = (v∘u)(x) = v(u(x))
Where u(x) = x⁴ and v(u(x)) = cos (u(x)) .
2. v'(u(x)) = −sin (u(x)) = −sin (x⁴)
3. u'(x) = 4x³
4. So f'(x) = v'(u(x)).u'(x) = −4x³ sin (x⁴)

◼ Using the results from parts (i) and (ii), we can write:
f'(x) =  −4 sin (x) cos³ (x) −4x³ sin (x⁴)

Solved example 21.31
Find the derivative of the function given by
f(x) = [g(x)]ⁿ
Solution:
1. f(x) = [g(x)]ⁿ = (v∘u)(x) = v(u(x))
Where u(x) = g(x) and v(u(x)) = (u(x))ⁿ .
2. v'(u(x)) = n(u(x))^(n-1) = n(g(x))^(n-1)
3. u'(x) = g'(x)
4. So f'(x) = v'(u(x)).u'(x) = n[g(x)]^(n-1) g'(x).

• The above result gives us an easy method to find the derivative in those cases where f(x) is the power of another function.

• For example, consider f(x) = cos²x.
Here, g(x) = cos x and n = 2.
So f'(x) = 2[cos x]. −sin x = −2 sin x cos x.

---

Link to a few more solved examples is given below:

Exercise 21.2

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We have completed a discussion on chain rule. In the next section, we will see derivatives of implicit functions.

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21.6 - The Chain Rule

In the previous section, we completed a discussion on differentiability. In this section, we will see derivatives of composite functions.

First we will see an example. It can be written in 3 steps:
1. We want the derivative of f, where f(x) = (3x+4)³.
2. Let us expand (3x+4)³. We get:
f(x) = 27x³ + 108 x² + 144 x + 64
3. This is a polynomial function. So we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} {f(x)}}    & {~=~}    &{\frac{d}{dx} {\left[(3x+4)^3 \right]}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{d}{dx} \left(27 x^3 + 108 x^2 + 144x + 64 \right)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{27(3)x^2 + 108 (2) x + 144 + 0}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{81x^2 + 216 x + 144 }    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{9 \left(9 x^2 + 24 x + 16 \right)}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{9 (3x + 4)^2}    \\
\end{array}$

---

Now we will see an easier method. The details about this new method can be written in 9 steps:

1. Consider the composite function f(x) = (v∘u)(x). It can also be written as f(x) = v(u(x)).
• We see that, the output of u is being used as the input for v.
2. The function in our example is: f(x) = (3x+4)³.
We can write it as the composite of two functions: f(x) = v(u(x))
    ♦ Where u(x) = 3x+4 and v(u(x)) = [u(x)]³.
• Then f(x) = v(u(x)) = v(3x+4) =  (3x+4)³.
3. Put u(x) = 3x+4 = t.
Then we get: f(x) = v(t) = t³ = (3x+4)³.
4. Recall the derivative that we obtained above, by considering it as a polynomial function.
• We obtained 9(3x+4)².
• So now we can write:
$\frac{d}{dx} {f(x)} \,=\, 9 (3x+4)^2 \,=\, \left[3 (3x+4)^2 \right].3\,=\, \left[3 t^2 \right].3$
5. Let us write the general form for 3t² and 3.
• We wrote: v(t) = t³.
   ♦ So 3t² is $\frac{dv}{dt}$
• Similarly we wrote: t = 3x+4.
   ♦ So 3 is $\frac{dt}{dx}$
6. Now the result in (4) becomes:
$\frac{d}{dx} {f(x)} \,=\, 9 (3x+4)^2 \,=\, \left[3 (3x+4)^2 \right].3\,=\, \frac{dv}{dt}.\frac{dt}{dx}$
• Taking the first and last terms, we get:
$\frac{df}{dx} \,=\, \frac{dv}{dt}.\frac{dt}{dx}$
7. The result obtained in the above step (6) is known as the chain rule.
• Let us write the 4 steps for applying this rule:
(i) We are given the composite function: f(x) = (v∘u)(x)
• This can be written as f(x) = v(u(x))
• So output of u is the input of v. That means, u can be considered as the inner function.
(ii) We put inner function u(x) = t
From this, we can find $\frac{dt}{dx}$
(iii) Write the outer function v in terms of t.
From this, we can find $\frac{dv}{dt}$
(iv) Finally we can calculate $\frac{df}{dx}$ as follows:
$\frac{df}{dx} \,=\, \frac{dv}{dt}.\frac{dt}{dx}$
8. Chain rule can be extended to composite of 3 functions.
Let us write the 5 steps:
(i) We are given the composite function: f = w∘u∘v
(ii) We put the inner most function v = t
From this, we can find $\frac{dt}{dx}$
(iii) The next inner function is u.
• Write u in terms of t. This gives u(t)
• Put u(t) = s. From this, we can find $\frac{ds}{dt}$
(iv) The outer function is w.
• Write w in terms of s. This gives w(s)
• From this, we can find $\frac{dw}{ds}$
(v) Finally we can calculate $\frac{df}{dx}$ as follows:
$\frac{df}{dx} \,=\, \frac{dw}{ds}.\frac{ds}{dt}.\frac{dt}{dx}$
9. In this way, we can extend the chain rule to composite of 4 or more functions.

Solved example 21.21
Find the derivative of the function given by f(x) = sin (x²)
Solution:
1. f(x) = sin (x²) = (v∘u)(x) = v(u(x))
Where u(x) = x² and v(u(x)) = sin (u(x)).
2. Put u(x) = x² = t.
So $\frac{dt}{dx} = 2x$
3. v(u(x)) = sin (u(x))
• So v(t) = sin t
⇒ $\frac{dv}{dt} = \cos t = \cos (x^2) $
4. $\frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} = \cos (x^2) . 2x = 2x \cos (x^2)$

Solved example 21.22
Find the derivative of the function given by f(x) = tan (2x+3)
Solution:
1. f(x) = tan (2x+3) = (v∘u)(x) = v(u(x))
Where u(x) = 2x+3 and v(u(x)) = tan (u(x)).
2. Put u(x) = 2x+3 = t.
So $\frac{dt}{dx} = 2$
3. v(u(x)) = tan (u(x))
So v(t) = tan t
⇒ $\frac{dv}{dt} = \sec^2 t = \sec^2 (2x+3)$
4. $\frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} = \sec^2 (2x+3) . 2 = 2 \sec^2 (2x+3)$

Solved example 21.23
Find the derivative of the function given by f(x) = sin (cos (x²))
Solution:
1. f(x) = sin (cos (x²)) = (w∘v∘u)(x) = (w∘v)(u(x))
Where u(x) = x²,  v(u(x)) = cos (u(x)) and w(v(u(x))) = sin (v(u(x))).
2. Put u(x) = x² = t.
So $\frac{dt}{dx} = 2x$
3. Put v(u(x)) = cos (u(x)) = cos t = s.
So $\frac{ds}{dt} = - \sin t = -\sin x^2$
4. w(v(u(x))) = sin (v(u(x)))
so w(s) = sin (s)
⇒ $\frac{dw}{ds} = \cos s = \cos (\cos t) = \cos (\cos x^2)$
5. $\frac{df}{dx} \,=\, \frac{dw}{ds}.\frac{ds}{dt}.\frac{dt}{dx}$
= cos (cos x²). −sin x². 2x
= −2x sin x². cos (cos x²)

Solved example 21.24
Find the derivative of the function given by
$f(x) = \sqrt{5x - 8}$
Solution:
1. $f(x) = \sqrt{5x - 8}$ = (v∘u)(x) = v(u(x))
Where u(x) = 5x−8 and v(u(x)) = $\sqrt{u(x)}$.
2. Put u(x) = 5x−8 = t.
So $\frac{dt}{dx} = 5$
3. v(u(x)) = $\sqrt{u(x)}$
So v(t) = $\sqrt{t}$
⇒ $\frac{dv}{dt} = \frac{1}{2} (t)^{- \frac{1}{2}} = \frac{1}{2 \sqrt{5x - 8}}$
4. $\frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} = \frac{1}{2 \sqrt{5x - 8}} . 5 = \frac{5}{2 \sqrt{5x - 8}}$

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We have seen the basics about chain rule. In the next section, we will see an alternate method for applying this rule.

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21.5 - Differentiability

In the previous section, we completed a discussion on continuous functions. In this section, we will see differentiability.

We have seen derivatives in class 11 (Details here). Let us recall some basic details. It can be written in 8 steps:
1. If f is a real function and c a point in it's domain, then the derivative of f at c is given by:
$f'(c) = \lim_{h\rightarrow 0} \left[\frac{f(c+h) - f(c)}{h} \right]$
2. We can find the derivative at a particular point.
• The derivative at a particular point c is denoted as f'(c).
3. We can also find the general form of the derivative.
• The general form is denoted as f'(x).
• In the general form, we can substitute c in the place of x. This will give the actual derivative at c.
4. f'(c) is the "slope of the tangent" of f at the point c.
5. f'(c) is also the "rate of change" of the output.
• For example, consider the function in which time (t) is the input and velocity (v) is the output. Then the derivative at t = 8 s will give the rate of change of velocity at the instant when the stop-watch reading is 8 s.
6. The process of finding the derivative of a function is called differentiation.
• Some text books use the following phrase:
Differentiate f(x) with respect to x.
   ♦ This means: we are being asked to find f'(x).
7. We saw three rules related to algebra of derivatives:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(u \pm v)'}    & {~=~}    &{u' \pm v'}    \\
{~\color{magenta}    2    }    &{{}}    &{(uv)'}    & {~=~}    &{u' v + u v'~\text{(Leibnitz rule or Product rule)}}    \\
{~\color{magenta}    3    }    &{{}}    &{\left(\frac{u}{v} \right)'}    & {~=~}    &{\frac{u' v \,-\, u v' }{v^2}~\text{Wherever}~v \ne 0~\text{(Quotient rule)}}    \\
\end{array}$

8. We saw some standard results also:

$\begin{array}{ll} {~\color{magenta}    1    }    &{\text{If}}    &{f(x)}    & {~=~}    &{x^n}    \\
{~\color{magenta}    {}    }    &{\text{   Then}}    &{f'(x)}    & {~=~}    &{n x^{n-1}}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    {}    &{{}}    \\
{~\color{magenta}    2    }    &{\text{If}}    &{f(x)}    & {~=~}    &{\sin x}    \\
{~\color{magenta}    {}    }    &{\text{   Then}}    &{f'(x)}    & {~=~}    &{\cos x}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    {}    &{{}}    \\
{~\color{magenta}    3    }    &{\text{If}}    &{f(x)}    & {~=~}    &{\cos x}    \\
{~\color{magenta}    {}    }    &{\text{   Then}}    &{f'(x)}    & {~=~}    &{- \sin x}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    {}    &{{}}    \\
{~\color{magenta}    4    }    &{\text{If}}    &{f(x)}    & {~=~}    &{\tan x}    \\
{~\color{magenta}    {}    }    &{\text{   Then}}    &{f'(x)}    & {~=~}    &{\sec^2 x}    \\
\end{array}$

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Now we will derive a more convenient expression for derivative. It can be done in 5 steps:

1. Consider the definition of derivative:
$f'(x)~=~\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}$
• We obtained this expression using fig.13.31 in section 13.13.
2. The above expression can be written in another convenient form by using fig.21.14 below:

Fig.21.14

• Point P has:
   ♦ x-coordinate c
   ♦ y-coordinate f(c)
• Point Q has:
   ♦ x-coordinate x
   ♦ y-coordinate f(x)
3. So base of the triangle PQR = PR = (x-c)
Also, altitude = QR = [f(x) − f(c)]
4. Then slope of the line PQ = $\frac{f(x) - f(c)}{x - c}$
• We want PQ to be the tangent at P. So we want Q to be very close to P. That is., we want x to be very close to c. We write this as x→c.
5. When x is very close to c, the slope of PQ is given by:
$\lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
• The slope in such a situation is:
the slope of tangent at P.
• Slope of tangent at P is the derivative at c.
• So we can write:
Derivative at c = $f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$

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Now we will see the "importance of continuity" in the process of differentiation. It can be written in steps:
1. We saw that, the derivative at c is given by:
$f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
2. The above expression can be rearranged as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(c)}    & {~=~}    &{\lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\lim_{x\rightarrow c}[f(x) - f(c)]}{\lim_{x\rightarrow c}[x - c]}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{\lim_{x\rightarrow c}[f(x)]~-~\lim_{x\rightarrow c}[f(c)]}{\lim_{x\rightarrow c}[x - c]}}    \\
\end{array}$                           

3. Consider the first term in the numerator:
$\lim_{x\rightarrow c}[f(x)]$
• It shows that, limit at c must exist.
4. We know that:
If limit at c is to exist, both left side and right side limits at c must exist, and they must be the same.
5. We have seen several examples where limit at a particular point does not exist. To cite one example, see fig.21.6 in section 21.2.
• In this example, we cannot find the derivative at x = 1 because, $\lim_{x\rightarrow 1}[f(x)]$ does not exist.    
6. While discussing the topic of continuity, we saw that:
If at a point c, both left side and right side limits exist and if they are equal, then the function is continuous at c.
7. Based on this condition for continuity, we can write the condition for differentiability.
• We can write:
If we want to find the derivative of a function f at the point c, then f must be continuous at c.
8. If it is possible to find the derivative of a function f at the point c, then we say that, f is differentiable at c.
• So we can write:
For differentiability at c, "continuity at c" is essential.
9. We have seen continuous functions. Such functions are continuous at every point in the domain.
• So we can write:
If the domain of a continuous function f, is [a,b], then f is differentiable at every point in [a,b]
• As in the case of continuity,
   ♦ At the point a, we need to consider only the right side limit.
   ♦ At the point b, we need to consider only the left side limit.

---

Now we will see a theorem which gives the relation between continuity and differentiability.
• The theorem states that:
If a function f is differentiable at a point c, then the function is continuous at c.

The proof can be written in 3 steps:
1. Consider the expression for derivative that we obtained above:
$f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$ 
2. Now consider the case when x ≠ c.
• We can write f(x) - f(c) as:
$f(x) - f(c) = {\frac{f(x) - f(c)}{x-c}}.(x-c)$
• Taking limits on both sides, we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\lim_{x\rightarrow c} \left[f(x) - f(c) \right]}    & {~=~}    &{\lim_{x\rightarrow c} \left[{\frac{f(x) - f(c)}{x-c}}.(x-c) \right]}    \\
{~\color{magenta}    2    }    &{\implies}    &{\lim_{x\rightarrow c} \left[f(x) \right] - \lim_{x\rightarrow c} \left[f(c) \right]}    & {~=~}    &{\lim_{x\rightarrow c} \left[{\frac{f(x) - f(c)}{x-c}} \right].\lim_{x\rightarrow c} \left[(x-c) \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{f'(c) . 0}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{\lim_{x\rightarrow c} \left[f(x) \right]}    & {~=~}    &{\lim_{x\rightarrow c} \left[f(c) \right]}    \\
{~\color{magenta}    6    }    &{\implies}    &{\lim_{x\rightarrow c} f(x)}    & {~=~}    &{f(c)}    \\
\end{array}$                           

3. We want to prove that, f is continuous at c. Recall that, we need to check two conditions to prove continuity.
(i) Limit exists at c.
(ii) $\lim_{x\rightarrow c} f(x) = f(c)$
• In the theorem, it is given that, f is differentiable at c. So it is obvious that, limit at c exists. Thus condition (i) is satisfied.
• From (3) we have: $\lim_{x\rightarrow c} f(x) = f(c)$.
Thus condition (ii) is also satisfied.
◼ Hence it is proved that, f is continuous at c.  

◼ As a corollary of the above theorem, we can write:
Every differentiable function is continuous.
(Corollary of a theorem, is a fact, which results directly from that theorem. The dictionary meaning can be seen here)

The converse of the above corollary is not true. That is., every continuous function is not differentiable. Let us see an example. It can be written in steps:
1. Consider the function f(x) = |x|.
• We know that, it is a continuous function. We want to prove that, it is not a differentiable function.
2. Let us try to find the derivative of this function at x = 0
3. We have the general equation:
$f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
• So the derivative at x = 0 is given by:
$f'(0) = \lim_{x\rightarrow 0} \left[\frac{|x| - f(0)}{x - 0} \right]$
4. Let us write the left side limit at x = 0:
$\lim_{x\rightarrow 0^{-}} \left[\frac{|x| - f(0)}{x - 0} \right]$
• As x approaches zero from the left side, the input will be a −ve value. Let it be −h.
• So the left side limit will be:
$\frac{|-h| - |0|}{(-h) - 0} ~=~\frac{h}{(-h)}~=~-1$ 
5. Let us write the right side limit at x = 0:
$\lim_{x\rightarrow 0^{+}} \left[\frac{|x| - f(0)}{x - 0} \right]$
• As x approaches zero from the right side, the input will be a +ve value. Let it be h.
• So the right side limit will be:
$\frac{|h| - |0|}{h - 0}~=~\frac{h}{h}~=~1$   
6. From (4) and (5), we see that:
Left side and right side limits are not the same. So the required limit does not exist at x = 0.
• Therefore, the derivative at x = 0 does not exist.
7. Based on this example, we can write:
All continuous functions are not differentiable.

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

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