NCERT Solutions for Exercise 5.4 Class 12 Maths Chapter 5 - Continuity and Differentiability

Question:2. Differentiate the following w.r.t. x:

$e^{\sin^{-1}x}$

Answer:

Given function is
$f(x)=e ^{\sin ^{-1}x}$
Let $g(x)={\sin ^{-1}x}$
Then,
$f(x)=e^{g(x)}$
Now, differentiation w.r.t. x
$f^{'}(x)=g^{'}(x).e^{g(x)}$ -(i)
$g(x) = \sin^{-1}x \Rightarrow g^{'}(x ) = \frac{1}{\sqrt{1-x^2}}$
Put this value in our equation (i)
$f^{'}(x) = \frac{1}{\sqrt{1-x^2}}.e^{\sin^{-1}x} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}$

Question:3. Differentiate the following w.r.t. x:

$e^{x^3}$

Answer:

Given function is
$f(x)=e ^{x^3}$
Let $g(x)=x^3$
Then,
$f(x)=e^{g(x)}$
Now, differentiation w.r.t. x
$f^{'}(x)=g^{'}(x).e^{g(x)}$ -(i)

$g(x) = x^3 \Rightarrow g'(x) = 3x^2$

Put this value in our equation (i)
$f^{'}(x) =3x^2.e^{x^3}$
Therefore, the answer is $3x^2.e^{x^3}$

Question:4. Differentiate the following w.r.t. x:

$\sin ( \tan ^ { -1} e ^{-x })$

Answer:

Given function is
$f(x)=\sin ( \tan ^ { -1} e ^{-x })$
Let's take $g(x ) = \tan^{-1}e^{-x}$
Now, our function reduces to
$f(x) = \sin(g(x))$
Now,
$f^{'}(x) = g^{'}(x)\cos(g(x))$ -(i)
And
$g(x)=\tan^{-1}e^{-x}\\\Rightarrow g^{'}(x) = \frac{d(\tan^{-1}e^{-x})}{dx}.\frac{d(e^{-x})}{dx}= \frac{1}{1+(e^{-x})^2}.-e^{-x} = \frac{-e^{-x}}{1+e^{-2x}}$
Put this value in our equation (i)
$f^{'}(x) =\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})$
Therefore, the answer is $\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})$

Question:5. Differentiate the following w.r.t. x:

$\log (\cos e ^x )$

Answer:

Given function is
$f(x)=\log (\cos e ^x )$
Let's take $g(x ) = \cos e^{x}$
Now, our function reduces to
$f(x) = \log(g(x))$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{g(x)}$ -(i)
And
$g(x)=\cos e^{x}\\\Rightarrow g^{'}(x) = \frac{d(\cos e^{x})}{dx}.\frac{d(e^{x})}{dx}= (-\sin e^x).e^{x} = -e^x.\sin e^x$
Put this value in our equation (i)
$f^{'}(x) =-e^x.\sin e^x.\frac{1}{\cos e^x} = -e^x.\tan e^x \ \ \ \ \ (\because \frac{\sin x}{\cos x}=\tan x)$
Therefore, the answer is $-e^x.\tan e^x,\ \ \ e^x\neq (2n+1)\frac{\pi}{2},\ \ n\in N$

Question:6. Differentiate the following w.r.t. x:

$e ^x + e ^{x^2} + .....e ^{x^5}$

Answer:

Given function is
$f(x)= e ^x + e ^{x^2} + .....e ^{x^5}$
Now, differentiation w.r.t. x is
$f^{'}(x)= \frac{d(e^x)}{dx}.\frac{d(x)}{dx}+\frac{d(e^{x^2})}{dx}.\frac{d(x^2)}{dx}+\frac{d(e^{x^3})}{dx}.\frac{d(x^3)}{dx}+\frac{d(e^{x^4})}{dx}.\frac{d(x^4)}{dx}+\frac{d(e^{x^5})}{dx}.\frac{d(x^5)}{dx}$
$=e^x.1+e^{x^2}.2x+e^{x^3}.3x^2+e^{x^4}.4x^3+e^{x^5}.5x^4$
$=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$
Therefore, answer is $e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$

Question:7. Differentiate the following w.r.t. x:

$\sqrt { e ^{ \sqrt x }} , x > 0$

Answer:

Given function is
$f(x)=\sqrt { e ^{ \sqrt x }}$
Lets take $g(x ) = \sqrt x$
Now, our function reduces to
$f(x) = \sqrt {e^{g(x)}}$
Now,
$f^{'}(x) = g^{'}(x) \cdot \frac{1}{2\sqrt{e^{g(x)}}} \cdot \frac{d\left(e^{g(x)}\right)}{dx} = g^{'}(x) \cdot \frac{1}{2\sqrt{e^{g(x)}}} \cdot e^{g(x)} = \frac{g^{'}(x) \cdot e^{g(x)}}{2\sqrt{e^{g(x)}}} = \frac{g^{'}(x) \cdot e^{\sqrt{x}}}{2\sqrt{e^{\sqrt{x}}}} \text{ -(i)}$
And
$g(x)=\sqrt x\\\Rightarrow g^{'}(x) = \frac{(\sqrt x)}{dx}=\frac{1}{2\sqrt x}$
Put this value in our equation (i)
$f^{'}(x) = \frac{e^{\sqrt{x}}}{2\sqrt{x} \cdot 2\sqrt{e^{\sqrt{x}}}} = \frac{e^{\sqrt{x}}}{4\sqrt{x e^{\sqrt{x}}}}$
Therefore, the answer is $\frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}.\ \ x>0$

Question:8 Differentiate the following w.r.t. x: $\log ( \log x ) , x > 1$

Answer:

Given function is
$f(x)=\log ( \log x )$
Lets take $g(x ) = \log x$
Now, our function reduces to
$f(x) = \log(g(x))$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{g(x)}$ -(i)
And
$g(x)=\log x\\\Rightarrow g^{'}(x) = \frac{1}{x}$
Put this value in our equation (i)
$f^{'}(x) =\frac{1}{x}.\frac{1}{\log x} = \frac{1}{x\log x}$
Therefore, the answer is $\frac{1}{x\log x}, \ \ x>1$

Question:9. Differentiate the following w.r.t. x:

$\frac{\cos x }{\log x} , x > 0$

Answer:

Given function is
$f(x)=\frac{\cos x }{\log x}$
We differentiate with the help of Quotient rule
$f^{'}(x)=\frac{\frac{d(\cos x)}{dx}.\log x-\cos x.\frac{(\log x)}{dx} }{(\log x)^2 }$
$=\frac{(-\sin x).\log x-\cos x.\frac{1}{x} }{(\log x)^2 } = \frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}$
Therefore, the answer is $\frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}$

Question:10. Differentiate the following w.r.t. x:

$\cos ( log x + e ^x ) , x > 0$

Answer:

Given function is
$f(x)=\cos ( log x + e ^x )$
Lets take $g(x) = ( log x + e ^x )$
Then , our function reduces to
$f(x) = \cos (g(x))$
Now, differentiation w.r.t. x is
$f'(x) = g'(x)(-\sin(g(x))) \tag{i}$
And
$g(x) = ( log x + e ^x )$
$g^{'}(x)= \frac{d(\log x)}{dx}+\frac{d(e^x)}{dx} = \frac{1}{x}+e^x$
Put this value in our equation (i)
$f^{'}(x) = -\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x)$
Therefore, the answer is $-\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x), x>0$