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

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

$\cos (\sin x)$

Answer:

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

$\sin (ax+b)$

Answer:

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

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

Answer:

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)
$\frac{df(k)}{dk}=\frac{d(\sec k)}{dk}=\sec k\tan k=\sec (\tan \sqrt{x})\tan (\tan \sqrt{x})$
$(\because k=\tan tandt=\sqrt{x})$
$\frac{df(t)}{dt}=\frac{d(\tan t)}{dt}={\sec }^{2}t={\sec }^2(\sqrt{x})(\because t=\sqrt{x})$
$\frac{dt}{dx}=\frac{d(\sqrt{x})}{dx}=\frac{1}{2\sqrt{x}}$
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

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

Answer:

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

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

Answer:

Given function is
$f(x)=\cos x^3.{\sin }^2(x^5)$
Differentitation w.r.t. x is
$f^{{}^{\prime }}(x)=g^{{}^{\prime }}(x).h(x)+g(x).h^{{}^{\prime }}(x)$
$g(x)=\cos x^{3}andh(x)=si{n}^2(x^5)$
Lets take $u=x^{3}andv=x^5$
Our functions become,
$\cos x^3=\cos u$ and ${\sin }^2(x^5)={\sin }^{2}v$
Now,
$g^{{}^{\prime }}(x)=\frac{d(g(x))}{dx}=\frac{d(g(u))}{du}.\frac{du}{dx}$ ( By chain rule)
$\frac{d(g(u))}{du}=\frac{d(\cos u)}{du}=-\sin u=-\sin x^3(\because u=x^3)$
$\frac{du}{dx}=\frac{d(x^3)}{dx}=3x^2$
$g^{{}^{\prime }}(x)=-\sin x^3.3x^2$ -(i)
Similarly,
$h^{{}^{\prime }}(x)=\frac{d(h(x))}{dx}=\frac{d(h(v))}{dv}.\frac{dv}{dx}$
$\frac{d(h(v))}{dv}=\frac{d({\sin }^{2}v)}{dv}=2\sin v\cos v=2\sin x^5\cos x^5(\because v=x^5)$

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

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

Answer:

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)
$\frac{d(f(k))}{dk}=\frac{d(2k)}{dk}=2$
$\frac{dk}{dt}=\frac{d(\sqrt{\cot t})}{dt}=\frac{1}{2\sqrt{cott}}.(-cose{c}^{2}t)=\frac{-cose{c}^2{x}^2}{2\sqrt{cot{x}^2}}(\because t=x^2)$
$\frac{dt}{dx}=\frac{d(x^2)}{dx}=2x$
So,
$\frac{d(f(k))}{dx}=2.\frac{-cose{c}^2{x}^2}{2\sqrt{cot{x}^2}}.2x=\frac{-2\sqrt{2}x}{{\sin }^2{x}^2\sqrt{\frac{2\sin x^2\cos x^2}{{\sin }^2{x}^2}}}$ ( Multiply and divide by $\sqrt{2}$ and multiply and divide $\sqrt{\cot x^2}$ by $\sqrt{\sin x^2}$

$(\because \cot x=\frac{\cos x}{\sin x}andcosecx=\frac{1}{\sin x})$
$=\frac{-2\sqrt{2}x}{\sin x^2\sqrt{\sin 2x^2}}(\because 2\sin x\cos x=\sin 2x)$
There, the answer is $\frac{-2\sqrt{2}x}{\sin x^2\sqrt{\sin 2x^2}}$

Question:8 Differentiate the functions with respect to x in

$\cos (\sqrt{x})$

Answer:

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$f(x)=|x-1|,xϵR$ is not differentiable at x = 1.

Answer:

Given function is

$f(x)=|x-1|,x\in \mathbb{R}$

We know that any function is differentiable when both

$\underset{h\to 0^{-}}{lim}\frac{f(c+h)-f(c)}{h}$ and $\underset{h\to 0^{+}}{lim}\frac{f(c+h)-f(c)}{h}$ are finite and equal.

The required condition for the function to be differentiable at $x=1$ is

$\underset{h\to 0^{-}}{lim}\frac{f(1+h)-f(1)}{h}=\underset{h\to 0^{+}}{lim}\frac{f(1+h)-f(1)}{h}$.

Now, the Left-hand limit of the function at $x=1$ is

$\underset{h\to 0^{-}}{lim}\frac{f(1+h)-f(1)}{h}=\underset{h\to 0^{-}}{lim}\frac{|1+h-1|-|1-1|}{h}=\underset{h\to 0^{-}}{lim}\frac{|h|-0}{h}$

$=\underset{h\to 0^{-}}{lim}\frac{-h}{h}=-1(\because h<0)$.

The Right-hand limit of the function at $x=1$ is

$\underset{h\to 0^{+}}{lim}\frac{f(1+h)-f(1)}{h}=\underset{h\to 0^{+}}{lim}\frac{|1+h-1|-|1-1|}{h}=\underset{h\to 0^{+}}{lim}\frac{|h|-0}{h}$

$=\underset{h\to 0^{+}}{lim}\frac{h}{h}=1$.

Now, it is clear that

$\text{R.H.L. at }x=1\ne \text{L.H.L. at }x=1$.
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.

Answer:

Given function is

$f(x)=\lfloor x\rfloor ,0<x<3$

We know that any function is differentiable when both

$\underset{h\to 0^{-}}{lim}\frac{f(c+h)-f(c)}{h}$ and $\underset{h\to 0^{+}}{lim}\frac{f(c+h)-f(c)}{h}$ are finite and equal.

The required condition for the function to be differentiable at $x=1$ is

$\underset{h\to 0^{-}}{lim}\frac{f(1+h)-f(1)}{h}=\underset{h\to 0^{+}}{lim}\frac{f(1+h)-f(1)}{h}$.

Now, the Left-hand limit of the function at $x=1$ is

$\underset{h\to 0^{-}}{lim}\frac{f(1+h)-f(1)}{h}=\underset{h\to 0^{-}}{lim}\frac{\lfloor 1+h\rfloor -\lfloor 1\rfloor }{h}=\underset{h\to 0^{-}}{lim}\frac{0-1}{h}$

$=\underset{h\to 0^{-}}{lim}\frac{-1}{h}=-\mathrm{\infty }(\because h<0\Rightarrow 1+h<1,\therefore \lfloor 1+h\rfloor =0)$.

The Right-hand limit of the function at $x=1$ is

$\underset{h\to 0^{+}}{lim}\frac{f(1+h)-f(1)}{h}=\underset{h\to 0^{+}}{lim}\frac{\lfloor 1+h\rfloor -\lfloor 1\rfloor }{h}=\underset{h\to 0^{+}}{lim}\frac{1-1}{h}$

$=\underset{h\to 0^{+}}{lim}\frac{0}{h}=0(\because h>0\Rightarrow 1+h>1,\therefore \lfloor 1+h\rfloor =1)$.

Now, it is clear that

$\text{R.H.L. at }x=1\ne \text{L.H.L. at }x=1$, and $\text{L.H.L.}$ is not finite as well.

Therefore, the function $f(x)=\lfloor x\rfloor$ is not differentiable at $x=1$.

Similarly, for $x=2$,

The required condition for the function to be differentiable at $x=2$ is

$\underset{h\to 0^{-}}{lim}\frac{f(2+h)-f(2)}{h}=\underset{h\to 0^{+}}{lim}\frac{f(2+h)-f(2)}{h}$.

Now, the Left-hand limit of the function at $x=2$ is

$\underset{h\to 0^{-}}{lim}\frac{f(2+h)-f(2)}{h}=\underset{h\to 0^{-}}{lim}\frac{\lfloor 2+h\rfloor -\lfloor 2\rfloor }{h}=\underset{h\to 0^{-}}{lim}\frac{1-2}{h}$

$=\underset{h\to 0^{-}}{lim}\frac{-1}{h}=-\mathrm{\infty }(\because h<0\Rightarrow 2+h<2,\therefore \lfloor 2+h\rfloor =1)$.

The Right-hand limit of the function at $x=2$ is

$\underset{h\to 0^{+}}{lim}\frac{f(2+h)-f(2)}{h}=\underset{h\to 0^{+}}{lim}\frac{\lfloor 2+h\rfloor -\lfloor 2\rfloor }{h}=\underset{h\to 0^{+}}{lim}\frac{2-2}{h}$

$=\underset{h\to 0^{+}}{lim}\frac{0}{h}=0(\because h>0\Rightarrow 2+h>2,\therefore \lfloor 2+h\rfloor =2)$.
Now, it is clear that
R.H.L. at x= 2 $\ne$ 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