Chapter 23 - Integrals

In the previous section, we completed a discussion on applications of derivatives. In this chapter, we will see Integrals.

• We began our discussion on derivatives in section 13.10.
• There we saw the equation for finding the distance traveled by a freely falling body as: s = 4.9t².
• Since the distance s depends on time t, we can say that, s is a function of t. We write this as: s = f (t) = 4.9t²
• Derivative of this function is given by: ${\frac{ds}{dt}\,=\,f'(t) \,=\,9.8t}$
• This derivative can be used to find the instantaneous velocity at any instant t. For example, at the instant when the stop-watch shows 8 s, the velocity will be: f '(8) = 9.8(8) = 78.4 m/s.
• Now let us think in reverse:
If we are given the derivative, can we find the original function?
This chapter tries to answer the above question.

The process of integration
• Suppose that, we are given a derivative f '(x).
   ♦ The process of finding the original function f(x) is known as integration.
   ♦ This process is also known as anti differentiation.
• In fact, integration is the inverse process of differentiation.
• The original function that we obtain by integration is called primitive.

---

Let us see some basic details about integration. It can be written in 5 steps:
1. First we will see three examples:
Example (i): We know that ${\frac{d}{dx}(\sin x)\,=\,\cos x}$
• That is., cos x is the derivative of sin x.
• This is same as: sin x is an anti derivative of cos x
   ♦ An anti derivative is also called an integral.       
Example (ii): We know that ${\frac{d}{dx}\left(\frac{x^3}{3} \right)\,=\,x^2}$
• That is., x² is the derivative of ${\frac{x^3}{3}}$.
• This is same as: ${\frac{x^3}{3}}$ is an integral of x².
Example (iii): We know that ${\frac{d}{dx}\left(e^x \right)\,=\,e^x}$
• That is., e^x is the derivative of e^x.
• This is same as: e^x is an integral of e^x.

2. Recall that, the derivative of a constant function is zero.
• Now consider the example (i) that we saw above.
The original function may be the sum of sin x and a constant function C (where C is any real number). Even then, we will get the same derivative:
${\frac{d}{dx}(\sin x\,+\,C)\,=\,\cos x \,+\, 0 \,=\, \cos x}$
So the integral is: sin x + C
• Similarly, consider the example (ii).
The original function may be the sum of ${\frac{x^3}{3}}$ and a constant function C (where C is any real number). Even then, we will get the same derivative:
${\frac{d}{dx}\left(\frac{x^3}{3} \,+\, C \right)\,=\,x^2 \,+\, 0 \,=\, x^2}$
So the integral is ${\frac{x^3}{3} \,+\, C}$
• Similarly, consider the example (iii).
The original function may be the sum of ${e^x}$ and a constant function C (where C is any real number). Even then, we will get the same derivative:
${\frac{d}{dx}(e^x \,+\, C )\,=\,e^x \,+\, 0 \,=\, e^x}$
So the integral is ${e^x \,+\, C}$

3. The constant C can be chosen arbitrarily from the set of real numbers. That means, C can be any random real number.
• So there are infinite possibilities for the integral. That means, there will be infinite integrals for a given derivative.
• For each of those infinite integrals, the derivative will be the same.
   ♦ The constant C is called arbitrary constant.
   ♦ The constant C is also called constant of integration.
• By varying C, we can get infinite anti derivatives (integrals) for any given derivative.

4. Since there are infinite integrals with the same derivative, we will write a general form. It can be explained in 4 steps:
(i) Suppose that, we are given a derivative.
• Recall that, derivative is also a function. So we will denote the given derivative as f(x).
(ii) Let the integral obtained by the integration of f(x), be F(x).
• Then we can write: $\frac{d}{dx} \left[F(x) \,+\, C \right]\,=\,f(x)$
   ♦ Where C is the constant of integration.
(iii) From the above expression, it is clear that:
By the integration of f(x), we will get F(x) + C
• This [F(x) + C] is a function.
   ♦ F(x) is unique.
   ♦ C can have infinite values.
(iv) Since C can have infinite values, there are infinite functions of the form [F(x) + C]
• We can write a set of all those functions as: {F+C, C ∈ R}
• This set is the general form of all integrals of f.

5. Now we will see an important property of integrals. It can be written in 6 step:
(i) Suppose that, we are given a function f.
• Pick any two integrals from among the infinite possible integrals. Let the two picked integrals be:
   ♦ F(x) + C₁
   ♦ F(x) + C₂
(ii) So we have two functions: F(x) + C₁ and F(x) + C₂
• Let us name them as g and h. So we have:
   ♦ g(x) = F(x) + C₁
   ♦ h(x) = F(x) + C₂
(iii) Now we can write the difference of g and h. We get:
g(x) − h(x) = [F(x) + C₁] − [F(x) + C₂] = [C₁ − C₂]
(iv) But [C₁ − C₂] is a constant. Let us call it C₃.
• So we can write: g(x) − h(x) = C₃
(v) Let us differentiate [g(x) − h(x)]. We get:
$\frac{d}{dx}[g(x) - h(x)]\,=\,\frac{d}{dx}[C_3]\,=\,0$
(vi) From (ii), we know that, g and h have the same derivative.
• So we can write:
If g and h are any two functions which have the same derivative, then the derivative of the "difference of g and h" will be zero.

---

• We have seen the basics of integration. In short, we can write:
We are given the derivative f(x). The process of finding the corresponding [F(x) + C] is called integration.

---

• We use the symbol $\int{f(x) \, dx}$ to denote the process of integration.
• When we see this symbol, we say:
Integration of f with respect to x.
• So we can write: $\int{f(x) \, dx}~=~F(x) \,+\,C$

---

Following table shows the meanings of various symbols/terms/phrases related to integration.

Table 23.1

---

• When we learned about derivatives, we saw many useful formulas. For example: $\frac{d}{dx} (\tan x) \,=\, \sec^2 x$.
• This formula can be used to write the corresponding formula for integral. We get: $\int{\sec^2 x \, dx}\,=\,\tan x \,+\, C$
• In this way, using the already known derivative formulas, we can make a list of integral formulas. A small list is given below:

$\begin{array}{ll} {~\color{magenta}    {}    }    &{{}}    &{\text{Derivatives}}    & {{}}    &{\text{Integrals}}    \\
{~\color{}    (i)    }    &{{}}    &{\frac{d}{dx}\left(\frac{x^{n+1}}{n+1} \right)~=~x^n}    & {\implies}    &{\int{x^n \, dx}~=~ \frac{x^{n+1}}{n+1}  \,+\, C}    \\
{~\color{}    {}    }    &{{}}    &{{}}    & {{}}    &{{}}    \\
{~\color{}    (ii)    }    &{{}}    &{\frac{d}{dx}\left(x \right)~=~1}    & {\implies}    &{\int{(1) \, dx}~=~ x \,+\, C}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{{}}    \\
{~\color{}    (iii)    }    &{{}}    &{\frac{d}{dx}\left(\sin x \right)~=~\cos x}    & {\implies}    &{\int{\cos x \, dx}~=~ \sin x \,+\, C}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{{}}    \\
{~\color{}    (iv)    }    &{{}}    &{\frac{d}{dx}\left(\cos^{-1} x \right)~=~ \frac{-1}{\sqrt{1 - x^2}}}    & {\implies}    &{\int{\frac{-1}{\sqrt{1 - x^2}} \, dx}~=~ \cos^{-1} x \,+\, C}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{{}}    \\
{~\color{}    (v)    }    &{{}}    &{\frac{d}{dx}\left(\log |x| \right)~=~\frac{1}{x}}    & {\implies}    &{\int{\frac{1}{x} \, dx}~=~ \log |x| \,+\, C}    \\
{~\color{}    {}    }    &{{}}    &{{}}    & {{}}    &{{}}    \\
\end{array}$

• The reader may add as many items as possible, to this list. We will be using the items in the list to solve complicated problems.

---

In the next section, we will see geometrical interpretation of indefinite integral.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

21.17 - Derivatives of Parametric Functions

In the previous section, we saw logarithmic differentiation. In this section, we will see parametric functions.

We have learned about projectile motion in our physics classes. An example in projectile motion, can be used to learn about parametric functions. It can be written in 6 steps:
1. An object is projected with a velocity of 30 ms⁻¹ at an angle of 60^o with the horizontal. Trace the path of the object. [g = 9.81 ms⁻²]

2. We know that:
• The point of projection is considered as the origin.
• The x-coordinate is the horizontal distance of the object from the origin.
• The y-coordinate is the vertical distance of the object from the origin.
• At the instant when the object is projected, the reading in the stop-watch will be: t = 0.

3. We know how to find the x and y coordinates:
• The x-coordinate is given by: $x = \rm{v_0 \cos \theta_0 t}$
        ✰ Eq.4.13 in section 4.10 of physics notes. 
• The y-coordinate is given by: $y = \rm{v_0 \sin \theta_0 t \,-\,\frac{1}{2}gt^2}$
        ✰ Eq.4.14 in section 4.10 of physics notes. 
• Where,
    ♦ v₀ is the initial velocity
    ♦ 𝜃₀ is the angle of projection.
    ♦ t is the time at which we want the coordinates.

4. We see that, both x and y coordinates are given in terms of t.
    ♦ So x is a function of a third variable ‘t’.
    ♦ y is also a function of the third variable ‘t’.
• We can write: x = f(t) and y = g(t).
• Such functions are called parametric functions.
    ♦ The third variable ‘t’ is the parameter.

5. Let us plot the various x and y positions.
• The table below gives the x and y coordinates:

(In this table, the last entry of the "time column" is obtained by finding the “time duration of the flight”) 
• The coordinates from the table, are plotted in the graph in fig.21.22 below:

Fig.21.22

• The pink smooth curve is drawn through the plotted points. Two sample plotted points are shown in the fig.
• The pink curve represents the path of the projected object. We obtained this path using a parametric function.

6. When we learned about projectile motion in the physics classes, we were not aware about parametric functions. There we plotted the pink curve by eliminating ‘t’ from the expressions for x and y. Such an elimination may not be possible in some complicated problems. For those complicated problems, parametric functions will help us to obtain the required results easily.

---

Now we have a basic idea about parametric functions. So we will learn the method to find the derivatives of such functions. It can be written in steps:
1. Consider the parametric equation:
x = f(t), y = g(t)

2. Assume that, by eliminating t, a direct relation F can be obtained between x and y. We can write it as: y = F(x).

3. Differentiating both sides w.r.t 't', we get:
$\rm{\frac{dy}{dt} ~=~\frac{d}{dt}(F(x))}$    

4. In the L.H.S. of (3), y is equal to g(t). So 't' is the variable. We can easily calculate $\rm{\frac{dy}{dt}}$

5. In the R.H.S. of (3), F(x) does not have t. This is because, F(x) is the direct relation between x and y. It is obtained by eliminating 't'.
• So we cannot find $\rm{\frac{d}{dt}(F(x))}$ directly. We need to use the chain rule. We get:
$\rm{\frac{d}{dt}(F(x))}~=~\frac{dF}{dx}.\frac{dx}{dt}$

6. So the equation in (3) becomes:
$\rm{\frac{dy}{dt} ~=~\frac{dF}{dx}.\frac{dx}{dt}}$
• Consider the R.H.S. of the above equation.
• From (2) we see that, F is same as y. So this equation becomes:
$\rm{\frac{dy}{dt} ~=~\frac{dy}{dx}.\frac{dx}{dt}}$
• From this we get:
$\rm{\frac{dy}{dx}~=~\frac{\frac{dy}{dt}}{\frac{dx}{dt}}}$

7. We can use the result in (6) above, to find $\rm{\frac{dy}{dx}}$, when x and y are given in terms of a third variable 't'.

---

Now we will see a solved example.

Solved example 21.55
$\rm{\text{Find}~\frac{dy}{dx}~\text{if}~x = a \cos \theta,~y = a \sin \theta}$.
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{a \cos \theta}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dx}{d \theta}}    & {~=~}    &{-a \sin \theta}    \\
{~\color{magenta}    3    }    &{{}}    &{y}    & {~=~}    &{a \sin \theta}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{dy}{d \theta}}    & {~=~}    &{a \cos \theta}    \\
{~\color{magenta}    5    }    &{\text{Therefore}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{a \cos \theta}{-a \sin \theta}}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{-\cot \theta}    \\
\end{array}$

---

In the next section, we will see a few more solved examples.

Previous

Contents


Next

Copyright©2024 Higher secondary mathematics.blogspot.com

21.16 - Logarithmic Differentiation

In the previous section, we saw the methods to find the derivatives of exponential and logarithmic functions. In this section, we will see logarithmic differentiation.

The basic details can be written in 3 steps:
1. Consider the function: $\rm{y=f(x)=[u(x)]^{v(x)}}$
    ♦ The base is a function of x.
    ♦ Exponent is also a function of x.

2. For differentiating this type of functions, we take logarithm on both sides. We get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log(y)}    & {~=~}    &{\log \left([u(x)]^{v(x)} \right)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\log(y)}    & {~=~}    &{v(x) \log [u(x)]}    \\
\end{array}$

3. Now we can apply the chain rule and obtain the derivative. This is shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log(y)}    & {~=~}    &{v(x) \log [u(x)]}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{d}{dx} \left(\log(y) \right)}    & {~=~}    &{\frac{d}{dx} \left(v(x) \log [u(x)] \right)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{1}{y} . \frac{d}{dx}(y)}    & {~=~}    &{\frac{d}{dx} \left(v(x) \right) . \log[u(x)]\,+\,v(x) . \frac{d}{dx} \left(\log[u(x)] \right)}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{1}{y} . \frac{dy}{dx}}    & {~=~}    &{v'(x) . \log[u(x)]\,+\,v(x) . \frac{1}{u(x)} \frac{d}{dx} \left(u(x) \right)}    \\
{~\color{magenta}    5    }    &{\implies}    &{\frac{1}{y} . \frac{dy}{dx}}    & {~=~}    &{v'(x) . \log[u(x)]\,+\,v(x) . \frac{1}{u(x)} u'(x)}    \\
{~\color{magenta}    6    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{y \left[v'(x) . \log[u(x)]\,+\,v(x) . \frac{1}{u(x)} u'(x) \right]}    \\
\end{array}$                           

◼ Remarks:
• 2(magenta color):
    ♦ We are taking the logarithm of y. That means, we are taking the logarithm of f(x). So f(x) must be +ve.
    ♦ Similarly, we are taking the logarithm of u(x). So u(x) must be +ve.

---

Let us see some solved examples:

Solved example 21.51
Find $\rm{\frac{dy}{dx}~\text{if}~y\,=\,\sqrt{\frac{(x-3)(x^2 + 4)}{3x^2 + 4x + 5}}}$
Solution:
1. Let us name the functions, so that, given y can be written in a short form:
    ♦ u(x) = x − 3
    ♦ v(x) = x²+4
    ♦ z(x) = 3x² + 4x + 5
• Then $\rm{y\,=\,\sqrt{\frac{u(x).v(x)}{z(x)}}}$ 

2. Taking logarithm on both sides, we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log(y)}    & {~=~}    &{\log \left(\sqrt{\frac{u(x).v(x)}{z(x)}} \right)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \log \left({\frac{u(x).v(x)}{z(x)}} \right)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left(\log u(x)\,+\,\log v(x) \,-\, \log z(x) \right)}    \\
\end{array}$                           

3. Now we can find the derivative as follows:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log(y)}    & {~=~}    &{\frac{1}{2} \left(\log u(x)\,+\,\log v(x) \,-\, \log z(x) \right)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{d}{dx} \left(\log(y) \right)}    & {~=~}    &{\frac{d}{dx} \left[\frac{1}{2} \left(\log u(x)\,+\,\log v(x) \,-\, \log z(x) \right) \right]}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{1}{y}.\frac{d}{dx} (y) }    & {~=~}    &{\frac{1}{2} \frac{d}{dx} \left[\log u(x)\,+\,\log v(x) \,-\, \log z(x) \right]}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{1}{y} . \frac{dy}{dx}}    & {~=~}    &{\frac{1}{2}  \left[\frac{d}{dx} \log u(x)\,+\,\frac{d}{dx} \log v(x) \,-\, \frac{d}{dx} \log z(x) \right]}    \\
{~\color{magenta}    5    }    &{\implies}    &{\frac{1}{y} . \frac{dy}{dx}}    & {~=~}    &{\frac{1}{2}  \left[\frac{1}{u(x)} . u'(x)\,+\,\frac{1}{v(x)} . v'(x) \,-\, \frac{1}{z(x)} . z'(x) \right]}    \\
{~\color{magenta}    6    }    &{\implies}    &{\frac{1}{y} . \frac{dy}{dx}}    & {~=~}    &{\frac{1}{2}  \left[\frac{1}{x-3} . (1)\,+\,\frac{1}{x^2 - 4} . (2x) \,-\, \frac{1}{3x^2 + 4x + 5} . (6x+4) \right]}    \\
{~\color{magenta}    7    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{y}{2} \left[\frac{1}{x-3} \,+\,\frac{2x}{x^2 - 4} \,-\, \frac{ 6x+4}{3x^2 + 4x + 5} \right]}    \\
{~\color{magenta}    8    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \sqrt{\frac{(x-3)(x^2 + 4)}{3x^2 + 4x + 5}} \left[\frac{1}{x-3} \,+\,\frac{2x}{x^2 - 4} \,-\, \frac{ 6x+4}{3x^2 + 4x + 5} \right]}    \\
\end{array}$

Solved example 21.52
Differentiate a^x w.r.t. x, where a is a positive constant.
Solution:
1. Let y = a^x.
2. Taking logarithm on both sides, we get:
log y = x log a.
3. Now we can find the derivative as follows:

Alternate method:
• First we will prove the identity: u = e^{log u}.
• Proof can be written in (ii) steps:
(i) Let v = e^{log u}.
(ii) Taking log on both sides, we get:
log v = log(e^{log u}) = log u.log e= log u.(1) = log u
⇒ log v = log u
⇒ v = u
⇒ e^{log u} = u

1. Let y = a^x.
2. $\rm{a^x~\text{is same as}~e^{\log a^x}}$
3. So we get: $\rm{y = e^{\log a^x}}$
4. Now we can find the derivative as follows:

 

Solved example 21.53
Differentiate x^{sin x} , x>0 w.r.t. x.
Solution:
1. Let y = x^{sin x}, x>0
2. Taking logarithm on both sides, we get:
log y = sin x log x.
3. Now we can find the derivative as follows:

Solved example 21.54
Find $\rm{\frac{dy}{dx}~\text{if}~y^x\,+\,x^y\,+\,x^x\,=\,a^b}$.
Solution:
1. Given that: y^x + x^y + x^x = a^b.
2. Let us write the L.H.S as the sum of three functions:
u(x) + v(x) + w(x) = a^b.
• Where:
   ♦ u(x) = y^x .
   ♦ v(x) = x^y .
   ♦ w(x) = x^x .
3. Now we get:

4. First we will find $\rm{\frac{du}{dx}}$:

5. Next we will find $\rm{\frac{dv}{dx}}$:

6. Finally we will find $\rm{\frac{dw}{dx}}$:

7. Substituting the above results in (3), we get:

---

The link below gives a few more solved examples:

Exercise 21.5

---

In the next section, we will see derivatives of functions in parametric forms.

Previous

Contents


Next

Copyright©2024 Higher secondary mathematics.blogspot.com

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'.

---

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.

---

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 $y=\frac{1}{x}$.

---

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

---

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

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

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:

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

---

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 = $f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$

---

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.

---

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

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com