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

You have already learned about the exponential and logarithmic functions in the previous classes. In the NCERT solutions for Class 12 Maths chapter 5 exercise 5.4, you will learn about the differentiation and logarithmic functions. The differentiation of simple exponential and logarithmic functions is very easy. In exercise 5.4 Class 12 Maths you will get questions related to the differentiation of composite functions containing the exponential and logarithmic functions.

As you learned the differentiation of composite functions in Class 12 Maths NCERT book exercise 5.2 already, you can use that concept to solve Class 12 Maths ch 5 ex 5.4 problems. Don't confuse with logarithmic differentiation, it is different from the differentiation of logarithmic function. You will about the logarithmic differentiation in the next exercise of Class 12 Maths NCERT syllabus. If you are looking for NCERT Solutions for other classes as well, click on the given link.

Also, see

• Continuity and Differentiability Exercise 5.1
• Continuity and Differentiability Exercise 5.2
• Continuity and Differentiability Exercise 5.3
• 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: 5.4

Question:1. Differentiate the following w.r.t. x:

$\frac{e ^x }{\sin x }$

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

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

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

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:

$e ^x^{ ^3}$

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:

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

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:

$\log (\cos e ^x )$

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:

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

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:

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

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,
$f^{'}(x) = g^{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.\frac{d({e^{g(x)}})}{dx} = g{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.{e^{g(x)}} = \frac{g^{'}(x).e^{g(x)}}{2.\sqrt{e^{g(x)}}} = \frac{g^{'}(x).e^{\sqrt x}}{2.\sqrt{e^{\sqrt x}}}$ -(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)
$f^{'}(x) =\frac{e^{\sqrt x}}{2\sqrt x.2.\sqrt{e^{\sqrt x}}} = \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}$
Therefore, the answer is $\frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}.\ \ x>0$

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

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:

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

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:

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

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))$ -(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 $-\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x), x>0$

## More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4:-

Class 12th Maths chapter 5 exercise 5.4 is all about the differentiation of composite functions containing exponential and logarithmic functions. There are two examples and few theorems given before the Class 12 Maths ch 5 ex 5.4 are related to that only. There are 10 questions in this exercise related to finding differentiation of exponential and logarithmic functions. You can easily solve these problems if you have done good practice of chain rule problems in exercise 5.2 of this chapter.

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

## Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4:-

• The differentiation of logarithmic and exponential is an important concept that will be useful not only in this maths but in all the other subjects also.
• NCERT solutions for Class 12 Maths chapter 5 exercise 5.4 are useful in learning this important concept of differentiation.
• Class 12 Maths chapter 5 exercise 5.4 solutions are designed following the guideline of CBSE.

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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