Continuity and Differentiability NCERT Solutions for Class 12 Maths Exercise 5.2

Differentiate the function with respect to x

Question 1
$\ sin (x^{2}+5)$
Solution
The given function is $f(x)=\ sin (x^{2}+5)$
We can see that $f$ is a composite function which can be written in the form of two composite function $u$ and $v$.
$u(x)=x^{2}+5$
$v(t)=\sin t$
$(vou)(x)=v(u(x))=v(x^{2}+5)=\sin (x^{2}+5)=f(x)$ put $t=(x^{2}+5)$
$\frac{\mathrm{d} v}{\mathrm{d} t}=\cos t$
And $\frac{\mathrm{d} t}{\mathrm{d} x}=2x$
By chain rule ,
$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$
$\frac{\mathrm{d} f}{\mathrm{d} x}=\cos t \times (2x)$
And we know $t=x^{2}+5$
Thus $\frac{\mathrm{d} f}{\mathrm{d} x}=(2x)\cos x^{2}+5$


Question 2
$\cos (\sin x)$
Solution
Let $f(x)=\cos (\sin x)$
We can see that $f$ is a composite function which can be written in the form of two composite function $u$ and $v$.
$u(x)=\sin x\; \; and\; \; v(t)=\cos t$
$(vou)(x)=v(u(x))=v(\sin x)=\cos (\sin x)=f(x)$
Put $t=u(x)=\sin x$
$therefore,\; \frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\cos t)}{\mathrm{d} t}=-\sin t = -\sin(\sin x)$
$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (\sin x)}{\mathrm{d} x}=\cos x$
By chain rule ,
$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$
$\frac{\mathrm{d} f}{\mathrm{d} x}=-\sin (\sin x)\times \ cos x=-\cos x\sin (\sin x)$

Question 3
$\sin (ax+b)$
Solution

Let $f(x)=\sin (ax+b)$
We can see that $f$ is a composite function which can be written in the form of two composite function $u$ and $v$.
$u(x)=ax+b\; \; and\; \; v(t)=\sin t$
$(vou)(x)=v(u(x))=v(ax+b)=\sin (ax+b)=f(x)$
Put $t=u(x)=ax+b$
$therefore,\; \frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\sin t)}{\mathrm{d} t}=\cos t = \cos(ax+b)$
$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (ax+b)}{\mathrm{d} x}=a$
By chain rule ,
$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$
$\frac{\mathrm{d} f}{\mathrm{d} x}=a\times \ cos (ax+b)$

Question 4
$\sec (\tan (\sqrt{x}))$
Solution

Let $f(x)=\sec (\tan (\sqrt{x}))$
We can see that $f$ is a composite function which can be written in the form of three composite function $u$,$v$ and $w$.
$u(x)=\sqrt{x}\; \; v(t)=\tan t\; \; and w(s)=\sec s$
$(wovou)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w[\tan( \sqrt{x})]=\sec (\tan (\sqrt{x}))=f(x)$
Put $s=v(t)=\tan t and t=u(x)=\sqrt{x}$
then ,$\frac{\mathrm{d} w}{\mathrm{d} s}=\frac{\mathrm{d} (\sec s)}{\mathrm{d} s}=\sec s\tan s$
Now $s=\tan t$ $=\sec(\tan t). \tan (\tan t)$
$=\sec(\tan \sqrt{x}). \tan (\tan \sqrt{x})$
$\frac{\mathrm{d} s}{\mathrm{d} t}=\frac{\mathrm{d} (\tan t)}{\mathrm{d} t}=\sec ^{2}t=\sec ^{2}\sqrt{t}$
$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} \sqrt{x}}{\mathrm{d} x}=\frac{\mathrm{d} (x^{\frac{1}{2}})}{\mathrm{d} x}=\frac{1}{2}.x^{\frac{1}{2}-1}=\frac{1}{2\sqrt{x}}$
By chain rule ,
$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} w}{\mathrm{d} s}\times \frac{\mathrm{d} s}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$
=$\sec (\tan\sqrt{x}).\tan (\tan\sqrt{x})\times \sec ^{2} \sqrt{x}\times \frac{1}{2\sqrt{x}}$
=$\frac{\sec ^{2}\sqrt{x}.\sec (\tan \sqrt{x}.\tan (\tan \sqrt{x}))}{2\sqrt{x}}$

Question 5
$\frac{\sin (ax+b)}{\cos(cx+b)}$
Solution
The given function is $f(x)=\frac{\sin (ax+b)}{\cos(cx+b)}= \frac{g(x)}{h(x)}$,
where $g(x)= \sin (ax+b) $ and $h(x)=\cos (cx+d)$
Differentiation of f(x) will be given by quotient formula
$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{g^{'}h-h^{'}g}{h^{2}}$
So we need to find the differentiation of the function g(x) and h(x)
We can use the same technique of chain rule here
Consider $g(x)=\sin (ax+b)$
Here $g$ is a composite function which can be written in the form of two composite function $u$ and $v$.
$u(x)=ax+b\; \; v(t)=\sin t$
$(vou)(x)=v(u(x))=v(ax+b)=\sin (ax+b)]=g(x)$
Put $t=u(x)=ax+b$
$\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\sin t)}{\mathrm{d} t}=\cos t=\cos (ax+b)$
$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (ax+b)}{\mathrm{d} x}=a$
By chain rule ,
$g^{‘}=\frac{\mathrm{d} g}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}=\cos (ax+b).a=a \cos (ax+b)$
Consider $h(x)=\cos (cx+d)$
Here $h$ is a composite function which can be written in the form of two composite function $p$ and $q$.
$p(x)=cx+d\; \; q(y)=\cos y$
$(qop)(x)=q(p(x))=q(cx+d)=\cos (cx+d)]=h(x)$
Put $y=p(x)=cx+d$
$\frac{\mathrm{d} q}{\mathrm{d} y}=\frac{\mathrm{d} (\cos y)}{\mathrm{d} y}=-\sin y=-\sin (cx+d)$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} (cx+d)}{\mathrm{d} x}=c$
By chain rule ,
$h^{‘}=\frac{\mathrm{d} h}{\mathrm{d} x}=\frac{\mathrm{d} q}{\mathrm{d} y}\times \frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (cx+d).c=-c \sin (cx+d)$
Therefore Differentiation of function f(x) will be given by
$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{a\cos (ax+b).\ cos (cx+d)-\sin (ax+b)(-\sin(cx+d))}{[\cos(cx+d)]^{2}}$
=$\frac{a\cos (ax+b)}{\ cos (cx+d)}+c\sin (ax+b).\frac{\sin (cx+d)}{\cos (cx+d)}\times \frac{1}{\cos (cx+d)}$
=$a\cos (ax+b)\ sec (cx+d)+c\sin (ax+b).\tan (cx+d).\sec (cx+d)$

Question 6
$\cos x^{3}.\sin ^{2}(x^{5})$
Solution
The given function is $f(x)=\cos x^{3}.\sin ^{2}(x^{5})$
By Multiplication(Product) Formula of differentiation
$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\left [ \cos x^{3}.\sin ^{2}(x^{5}) \right ]$
$=\sin ^{2}(x^{5})\times \frac{\mathrm{d} }{\mathrm{d} x}\cos x^{3}+\cos x^{3}\times \frac{\mathrm{d} }{\mathrm{d} x}\sin ^{2}(x^{5})$
=$\sin ^{2}(x^{5})\times (-\sin x^{3})\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{3})+\cos x^{3}\times 2\sin (x^{5})\times \frac{\mathrm{d} }{\mathrm{d} x}\left [ \sin (x^{5}) \right ]$
$=-\sin x^{3}.\sin ^{2}(x^{5}) \times 3x^{2}+2\sin (x^{5}).\cos x^{3}.\cos x^{5}\frac{\mathrm{d} }{\mathrm{d} x} (x^{5})$
$=-\sin x^{3}.\sin ^{2}(x^{5}) \times 3x^{2}+2\sin (x^{5}).\cos x^{3}.\cos x^{5}\times 5x^{4}$
$=10x^{4}\sin x^{5}.\cos x^{3}.\cos x^{5}\times -3x^{2}\sin x^{3}.\sin ^{2}(x^{5} )$


Question 7
$2\sqrt{\cot (x^{2})}$
Solution
The given function is $2\sqrt{\cot (x^{2})}$
$\frac{\mathrm{d} }{\mathrm{d} x}2\sqrt{\cot (x^{2})}$ $=2.\frac{1}{2\sqrt{\cot (x^{2})}}\times \frac{\mathrm{d} }{\mathrm{d} x}\left [ \cot (x^{2}) \right ]$ ( We use the chain rule here)
$=\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}}\times -cosec^{2} (x^{2})\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{2})$ ( We use the chain rule here)
$=-\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}}\times \frac{1}{\sin ^{2(x^{2})}}\times(2x)$
$=\frac{-2x}{\sqrt{\cos (x^{2})}\sqrt{\sin (x^{2})}\sin (x^{2})}$
$=\frac{-2\sqrt{2}x}{\sin (x^{2})\sqrt{\sin 2(x^{2})}}$

Question 8
$\cos\sqrt{x}$
Solution
The given function $f(x)$ is $\cos\sqrt{x}$.
We can see that $f$ is a composite function which can be written in the form of two composite function $u$ and $v$.
$u(x)=\sqrt{x}$
And $v(t)=\cos t$
$(vou)(x)=v(u(x))$
$=v(\sqrt{x})$
$=\cos (\sqrt{x})$
put $t=u(x)=\sqrt{x}$
Then, $\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{x})=\frac{\mathrm{d} }{\mathrm{d} x}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}$
$=\frac{1}{2\sqrt{x}}$
And, $\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} }{\mathrm{d} t}(\cos t)=-\sin t$
$=-\sin (\sqrt{x})$
By chain rule we have,
$\frac{\mathrm{d} v}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}.\frac{\mathrm{d} t}{\mathrm{d} x}$.
$=-\sin (\sqrt{x}).\frac{1}{2\sqrt{x}}$
$=-\frac{1}{2\sqrt{x}}\sin (\sqrt{x})$
$=-\frac{\sin (\sqrt{x})}{2\sqrt{x}}$

Question 9
Prove that the function f given by $f(x)=\left | x-1 \right |,x\in \mathbb{R}$, is not differentiable at $x=1$
Solution
The given function is $f(x)=\left | x-1 \right |,x\in \mathbb{R}$.
It is known that a function $f$ is differentiable at a point $x=c$ in its domain if the right hand limit and the left hand limit are finite and equal.
To check the differentiability of the given function at x=1,
The right hand and the left hand limits where x=c are
$\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}$ and $\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}$
Considering the right hand limit of the given function at x=1
$\lim_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}$
$=\lim_{h \to 0^{+}}\frac{\left | 1+h-1 \right |-\left | 1-1 \right |}{h}$
$=\lim_{h \to 0^{+}}\frac{\left | h \right |-\left | 0 \right |}{h}$
$=\lim_{h \to 0^{+}}\frac{h}{h}$
$=1$
Considering the left hand limit of the given function at x=1
$\lim_{h \to 0^{-}}\frac{f(1+h)-f(1)}{h}$
$=\lim_{h \to 0^{-}}\frac{\left | 1+h-1 \right |-\left | 1-1 \right |}{h}$
$=\lim_{h \to 0^{-}}\frac{\left | h \right |-\left | 0 \right |}{h}$
$=\lim_{h \to 0^{-}}\frac{-h}{h}$
$=-1$
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1

Question 10
Prove that the greatest integer function defined by $f(x)=\left [ x \right ], 0<x<3$ is not differentiable at x = 1 and x = 2.
Solution

The function f is $f(x)=\left [ x \right ], 0<x<3$
It is known that a function f is differentiable at a point $x=c$ in its domain if both the left hand and the left hand limit are equal
$\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}$ and $\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}$ are finite and equal.
To check the differentiability of the given function at $x=1$, consider the right hand limit of f at $x=1$
$\lim_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}$
$\lim_{h \to 0^{+}}\frac{\left [ 1+h \right ]-\left [ 1 \right ]}{h}$
$=\lim_{h \to 0^{+}}\frac{1-1}{h}$
$=\lim_{h \to 0^{+}}\frac{1-1}{h}=\lim_{h \to 0^{+}}(0)=0$
Now consider the left hand limit of f at $x=1$
$\lim_{h \to 0^{-}}\frac{f(1+h)-f(1)}{h}$
$\lim_{h \to 0^{-}}\frac{\left [ 1+h \right ]-\left [ 1 \right ]}{h}$
$=\lim_{h \to 0^{-}}\frac{0-1}{h}$
$=\lim_{h \to 0^{-}}\frac{-1}{h}=\infty $
Since left hand and the right hand limit of f at x=1 are not equal, f is not differentiable at x=1.
Now to check the differentiability of the given function at x=2,
consider the left hand limit at x=2.
$\lim_{h \to 0^{-}}\frac{f(2+h)-f(2)}{h}$
$\lim_{h \to 0^{-}}\frac{\left [ 2+h \right ]-\left [ 2 \right ]}{h}$
$=\lim_{h \to 0^{-}}\frac{1-2}{h}$
$=\lim_{h \to 0^{-}}\frac{-1}{h}=\infty $
Now consider the right hand limit of f at $x=2$
$\lim_{h \to 0^{+}}\frac{f(2+h)-f(2)}{h}$
$\lim_{h \to 0^{+}}\frac{\left [ 2+h \right ]-\left [ 2 \right ]}{h}$
$=\lim_{h \to 0^{+}}\frac{2-2}{h}$
$=\lim_{h \to 0^{+}}\frac{0}{h}=\lim_{h \to 0^{+}}(0)=0$
Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2

 

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• Notes
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• NCERT Solutions & Assignments
  • NCERT Solution for Continuity And differentiability Exercise 5.1
  • NCERT Solution for Continuity And differentiability Exercise 5.2
  • NCERT Solution for Continuity And differentiability Exercise 5.3
  • NCERT Solution for Continuity And differentiability Exercise 5.4
  • Continuity and Differentiability class 12 Important questions

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