NCERT Solutions for Exercise 5.2 Class 12 Maths Chapter 5 - Continuity and Differentiability

In the NCERT syllabus Class 11 Maths, you have already learned about the derivatives of real value functions. Differentiation is defined as the process of finding the derivative of a function. In this article, you will get NCERT solutions for Class 12 Maths chapter 5 Exercise 5.2. This NCERT book exercise consists of questions related to finding derivatives of different types of functions.

Mainly derivatives of composite functions using chain rule are covered in the Class 12 Maths ch 5 ex 5.2. The chain rule is a very important concept to find the derivative of a composite function of two differentiable functions using derivatives of these functions. Solve problems from exercise 5.2 Class 12 Maths to get an understanding of the chain rule. Also, you can try to prove this theorem to get conceptual clarity. If you looking for NCERT Solutions, you can click on the given link.

Also, see

• Continuity and Differentiability Exercise 5.1
• Continuity and Differentiability Exercise 5.3
• Continuity and Differentiability Exercise 5.4
• Continuity and Differentiability Exercise 5.5
• Continuity and Differentiability Exercise 5.6
• Continuity and Differentiability Exercise 5.7
• Continuity and Differentiability Exercise 5.8
• Continuity and Differentiability Miscellaneous Exercise

Continuity and Differentiability Exercise: 5.2

Question:1.Differentiate the functions with respect to x in

$\sin (x^2 +5 )$

Given function is
$f(x)=\sin (x^2 +5 )$
when we differentiate it w.r.t. x.
Lets take $t = x^2+5$ . then,
$f(t) = \sin t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (x^2+5)$
$\frac{dt}{dx} = \frac{d(x^2+5 )}{dx} = 2x$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (x^2+5).2x$
Therefore, the answer is $2x \cos (x^2+5)$

Question:2. Differentiate the functions with respect to x in

$\cos ( \sin x )$

Given function is
$f(x)= \cos ( \sin x )$
Lets take $t = \sin x$ then,
$f(t) = \cos t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ ( By chain rule)
$\frac{df(t)}{dt} = \frac{d(\cos t)}{dt} = -\sin t = -\sin (\sin x)$
$\frac{dt}{dx} = \frac{d(\sin x)}{dt} = \cos x$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = -\sin(\sin x).\cos x$
Therefore, the answer is $-\sin(\sin x).\cos x$

Question:3. Differentiate the functions with respect to x in

$\sin (ax +b )$

Given function is
$f(x) = \sin (ax +b )$
when we differentiate it w.r.t. x.
Lets take $t = ax+b$ . then,
$f(t) = \sin t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (ax+b)$
$\frac{dt}{dx} = \frac{d(ax+b )}{dx} = a$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (ax+b).a$
Therefore, the answer is $a \cos (ax+b)$

Question:4. Differentiate the functions with respect to x in

$\sec (\tan (\sqrt x) )$

Given function is
$f(x)=\sec (\tan (\sqrt x) )$
when we differentiate it w.r.t. x.
Lets take $t = \sqrt x$ . then,
$f(t) = \sec (\tan t)$
take $\tan t = k$. then,
$f(k) = \sec k$
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(k)}{dk} = \frac{d(\sec k )}{dk} = \sec k \tan k = \sec(\tan\sqrt x)\tan(\tan\sqrt x)$
$(\because k = \tan t \ and \ t = \sqrt x)$
$\frac{df(t)}{dt} = \frac{d(\tan t )}{dt} = \sec^2 t =\sec^2 (\sqrt x) \ \ \ \ \ \ (\because t = \sqrt x)$
$\frac{dt}{dx} = \frac{d(\sqrt x)}{dx} = \frac{1}{2\sqrt x}$
Now,
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx} =\sec(\tan \sqrt x)\tan(\tan \sqrt x).\sec^2 (\sqrt x) . \frac{1}{2\sqrt x}$
Therefore, the answer is $\frac{\sec(\tan \sqrt x).\tan(\tan \sqrt x).\sec^2 (\sqrt x)}{2\sqrt x}$

Question:5. Differentiate the functions with respect to x in

$\frac{\sin (ax +b )}{\cos (cx + d)}$

Given function is
$f(x) = \frac{\sin (ax +b )}{\cos (cx + d)} = \frac{g(x)}{h(x)}$
We know that,
$f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)}$
$g(x) = \sin(ax+b)$ and $h(x) = \cos(cx+d)$
Lets take $u = (ax+b) \ and \ v = (cx+d)$
Then,
$\sin (ax+b) = \sin u \ and \ \cos(cx+d) = \cos c$
$g^{'}(x)=\frac{d(g(x))}{dx} = \frac{d(g(x))}{du}.\frac{du}{dx}$ (By chain rule)
$\frac{d(g(x))}{du} = \frac{d(\sin u)}{du} = \cos u = \cos(ax+b) \ \ \ \ \ \ \ \ \ (\because u = ax +b)$
$\frac{du}{dx} = \frac{d(ax+b)}{dx} = a$
$g^{'}(x)=a\cos (ax+b)$ -(i)
Similarly,
$h^{'}(x)=\frac{d(h(x))}{dx} = \frac{d(h(x))}{dv}.\frac{dv}{dx}$
$\frac{d(h(x))}{dv}= \frac{d(\cos v)}{dv} = -\sin v = -\sin (cx+d) \ \ \ \ \ \ \ (\because v = (cx+d))$
$\frac{dv}{dx}= \frac{d(cx+d)}{dv} = c$
$h^{'}(x)=-c\sin(cx+d)$ -(ii)
Now, put (i) and (ii) in
$f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)} = \frac{a\cos(ax+b).\cos(cx+d)-\sin(ax+b).(-c.\sin(cx+d))}{\cos^2(cx+d)}$
$= \frac{a\cos(ax+b).\cos(cx+d)}{\cos^2(cx+d)}+\frac{\sin(ax+b).c.\sin(cx+d)}{\cos^2(cx+d)}$
$= a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)$
Therefore, the answer is $a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)$

Question:6. Differentiate the functions with respect to x in

$\cos x^3 . \sin ^ 2 ( x ^5 )$

Given function is
$f(x)=\cos x^3 . \sin ^ 2 ( x ^5 )$
Differentitation w.r.t. x is
$f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x)$
$g(x) = \cos x^3 \ and \ h(x) = sin^2(x^5)$
Lets take $u = x^3 \ and \ v = x^5$
Our functions become,
$\cos x^3 = \cos u$ and $\sin^2(x^5) = \sin^2v$
Now,
$g^{'}(x) = \frac{d(g(x))}{dx} =\frac{d(g(u))}{du}.\frac{du}{dx}$ ( By chain rule)
$\frac{d(g(u))}{du} = \frac{d(\cos u)}{du} = -\sin u =- \sin x^3 \ \ \ \ (\because u = x^3)$
$\frac{du}{dx} = \frac{d(x^3)}{dx} = 3x^2$
$g^{'}(x) = -\sin x^3.3x^2$ -(i)
Similarly,
$h^{'}(x) = \frac{d(h(x))}{dx} =\frac{d(h(v))}{dv}.\frac{dv}{dx}$
$\frac{d(h(v))}{dv}= \frac{d(\sin^2v)}{dv} =2\sin v \cos v =2\sin x^5\cos x^5 \ \ \ (\because v = x^5)$

$\frac{dv}{dx} = \frac{d(x^5)}{dx} = 5x^4$
$h^{'}(x) = 2\sin x^5\cos x^5.5x^4 = 10x^4\sin x^5\cos x^5$ -(ii)
Put (i) and (ii) in
$f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x) = -3x^2\sin x^3.\sin^2 x^5+\cos x^3.10x^4\sin x^5\cos x^5$
Therefore, the answer is $10x^4\sin x^5\cos x^5.\cos x^3 -3x^2\sin x^3.\sin^2 x^5$

Question:7. Differentiate the functions with respect to x in

$2 \sqrt { \cot ( x^2 )}$

Give function is
$f(x)=2 \sqrt { \cot ( x^2 )}$
Let's take $t = x^2$
$f(t) = 2\sqrt{\cot t}$
Now, take $\cot t = k^2$
$f(k) = 2k$
Differentiation w.r.t. x
$\frac{d(f(k))}{dx} = \frac{d(f(k))}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$ -(By chain rule)
$\frac{d(f(k))}{dk} = \frac{d(2k)}{dk} = 2$
$\frac{dk}{dt} = \frac{d(\sqrt{\cot t})}{dt} = \frac{1}{2\sqrt{cot t}}.(-cosec^2 t) = \frac{-cosec^2 x^2}{2\sqrt{cot x^2}} \ \ \ (\because t = x^2)$
$\frac{dt}{dx} = \frac{d(x^2)}{dx} = 2x$
So,
$\frac{d(f(k))}{dx} = 2.\frac{-cosec^2 x^2}{2\sqrt{cot x^2}}.2x = \frac{-2\sqrt2x}{\sin^2x^2\sqrt{\frac{2\sin x^2\cos x^2}{\sin^2x^2}} }$ ( Multiply and divide by $\sqrt 2$ and multiply and divide $\sqrt {\cot x^2}$ by $\sqrt{\sin x^2$)
$(\because \cot x = \frac{\cos x}{\sin x} \ and \ cosec x = \frac{1}{\sin x } )$
$=\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}} \ \ \ \ (\because 2\sin x\cos x=\sin2x)$
There, the answer is $\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}$

Question:8 Differentiate the functions with respect to x in

$\cos ( \sqrt x )$

Let us assume : $y\ =\ \cos ( \sqrt x )$

Differentiating y with respect to x, we get :

$\frac{dy}{dx}\ =\ \frac{d(\cos ( \sqrt x ))}{dx}$

or $=\ - \sin \sqrt{x}.\frac{d( \sqrt x )}{dx}$

or $=\ \frac{- \sin \sqrt{x}}{2\sqrt{x}}$

Question:9. Prove that the function f given by$f (x) = |x-1 | , x \epsilon R$ is not differentiable at x = 1.

Given function is
$f (x) = |x-1 | , x \epsilon R$
We know that any function is differentiable when both
$\lim_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h}$ and $\lim_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h}$ are finite and equal
Required condition for function to be differential at x = 1 is

$\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}$
Now, Left-hand limit of a function at x = 1 is
$\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h} = \lim_{h\rightarrow 0^-}\frac{|h|-0}{h}$
$= \lim_{h\rightarrow 0^-}\frac{-h}{h} = -1 \ \ \ \ (\because h < 0)$
Right-hand limit of a function at x = 1 is
$\lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h} = \lim_{h\rightarrow 0^+}\frac{|h|-0}{h}$
$=\lim_{h\rightarrow 0^-}\frac{h}{h} = 1$
Now, it is clear that
R.H.L. at x= 1 $\neq$ L.H.L. at x= 1
Therefore, function $f (x) = |x-1 |$ is not differentiable at x = 1

Question:10. Prove that the greatest integer function defined by $f (x) = [x] , 0 < x < 3$ is not differentiable at

x = 1 and x = 2.

Given function is
$f (x) = [x] , 0 < x < 3$
We know that any function is differentiable when both
$\lim_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h}$ and $\lim_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h}$ are finite and equal
Required condition for function to be differential at x = 1 is

$\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}$
Now, Left-hand limit of the function at x = 1 is
$\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^-}\frac{[1+h]-[1]}{h} = \lim_{h\rightarrow 0^-}\frac{0-1}{h}$
$=\lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ \ (\because h < 0 \Rightarrow 1+h<1, \therefore [1+h] =0)$
Right-hand limit of the function at x = 1 is
$\lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{[1+h]-[1]}{h} = \lim_{h\rightarrow 0^+}\frac{1-1}{h}$
$=\lim_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \Rightarrow 1+h>1, \therefore [1+h] =1)$
Now, it is clear that
R.H.L. at x= 1 $\neq$ L.H.L. at x= 1 and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 1
Similary, for x = 2
Required condition for function to be differential at x = 2 is

$\lim_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h}$
Now, Left-hand limit of the function at x = 2 is
$\lim_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^-}\frac{[2+h]-[2]}{h} = \lim_{h\rightarrow 0^-}\frac{1-2}{h}$
$=\lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ (\because h < 0 \Rightarrow 2+h<2, \therefore [2+h] =1)$
Right-hand limit of the function at x = 1 is
$\lim_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^+}\frac{[2+h]-[2]}{h} = \lim_{h\rightarrow 0^+}\frac{2-2}{h}$
$=\lim_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \Rightarrow 2+h>2, \therefore [2+h] =2)$
Now, it is clear that
R.H.L. at x= 2 $\neq$ L.H.L. at x= 2 and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 2

More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2:-

Class 12 Maths chapter 5 exercise 5.2 solutions is all about using the chain rule to find derivatives of composite functions. There are a total of 10 questions out of which 8 questions are from chain rule in Class 12 Maths ch 5 ex 5.2. There are 3 examples given before this exercise in the NCERT textbook, that you can solve to understand this concept fully. This concept is very easy as well useful for composite functions. You will find the use of this concept not only in this exercise but in physics, chemistry, and in other chapters also.

Also Read| Continuity and Differentiability Class 12th Chapter 5 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2:-

• NCERT Solutions for Class 12 Maths Chapter 5 exercise 5.2 is useful in solving this exercise as well as other subjects for finding derivatives of composite functions.
• There are other methods to find the derivative of a composite function, but the chain rule is more simple to use.
• The last two questions in exercise 5.2 Class 12 Maths are related to checking the differentiability of given functions
• You can use Class 12th Maths chapter 5 exercise 5.2 for quick revision of chain rule.

Also see-

• NCERT Solutions for Class 12 Maths Chapter 5

• NCERT Exemplar Solutions Class 12 Maths Chapter 5

NCERT Solutions of Class 12 Subject Wise

• NCERT Solutions for Class 12 Maths

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Chemistry

• NCERT Solutions for Class 12 Biology

Subject Wise NCERT Exampler Solutions

• NCERT Exemplar Solutions for Class 12th Maths

• NCERT Exemplar Solutions for Class 12th Physics

• NCERT Exemplar Solutions for Class 12th Chemistry

• NCERT Exemplar Solutions for Class 12th Biology

Happy learning!!!

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