# NCERT Solutions for Exercise 5.5 Class 12 Maths Chapter 5 - Continuity and Differentiability

In the previous exercise , you have already learned about the differentiation of logarithmic and exponential functions. In exercise 5.5 Class 12 Maths, you will learn about logarithmic differentiation which is not the same as differentiation of logarithmic functions. This trick of differentiation is used for the differentiation of functions raised to the power of functions or variables. In NCERT solutions for Class 12 Maths chapter 5 exercise 5.5, you will learn that logarithmic differentiation relies on the property of log and chain rule that you have already learned. If you have a good command of the chain rule of differentiation, you can easily solve these problems even without knowing about the logarithmic differentiation. You can go through these Class 12th Maths chapter 5 exercise 5.5 to get in-depth knowledge of this concept. Also, if you are looking for NCERT Solutions at one place, click on the given link above.

Also, see

• Continuity and Differentiability Exercise 5.1
• Continuity and Differentiability Exercise 5.2
• Continuity and Differentiability Exercise 5.3
• Continuity and Differentiability Exercise 5.4
• 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: 5.5

Question:1 Differentiate the functions w.r.t. x. $\cos x . \cos 2x .\cos 3x$

Given function is
$y=\cos x . \cos 2x .\cos 3x$
Now, take log on both sides
$\log y=\log (\cos x . \cos 2x .\cos 3x)\\ \log y = \log \cos x + \log \cos 2x + \log \cos 3x$
Now, differentiation w.r.t. x
$\log y=\log (\cos x . \cos 2x .\cos 3x)\\ \frac{d(\log y )}{dx} = \frac{\log \cos x}{dx} + \frac{\log \cos 2x}{dx} + \frac{\log \cos 3x}{dx}\\ \frac{1}{y}.\frac{dy}{dx} = (-\sin x)\frac{1}{\cos x}+(-2\sin 2x)\frac{1}{\cos 2x}+(-3\sin3x).\frac{1}{\cos3x}\\ \frac{1}{y}\frac{dy}{dx} = -(\tan x+\tan 2x+\tan 3x) \ \ \ \ \ \ (\because \frac{\sin x }{\cos x} =\tan x)\\ \frac{dy}{dx}=-y(\tan x+\tan 2x+\tan 3x)\\ \frac{dy}{dx}= -\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)$
There, the answer is $-\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)$

### Question:2. Differentiate the functions w.r.t. x.

$\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$

Given function is
$y=\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$
Take log on both the sides
$\log y=\frac{1}{2}\log\left ( \frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)} \right )\\ \log y = \frac{1}{2} (\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)-\log(x-5))\\$
Now, differentiation w.r.t. x is
$\frac{d(\log y)}{dx} = \frac{1}{2} (\frac{d(\log(x-1))}{dx}+\frac{d(\log(x-2))}{dx}-\frac{d(\log(x-3))}{dx}-\frac{d(\log(x-4))}{dx}-\\$$\frac{d(\log(x-5))}{dx})$
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\\ \frac{dy}{dx}=y\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\\ \frac{dy}{dx} = \frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$
Therefore, the answer is $\frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$

### Question:3 Differentiate the functions w.r.t. x. $(\log x ) ^{\cos x}$

Given function is
$y=(\log x ) ^{\cos x}$
take log on both the sides
$\log y=\cos x\log (\log x )$
Now, differentiation w.r.t x is
$\frac{d(\log y)}{dx} = \frac{d(\cos x\log(\log x))}{dx}\\ \frac{1}{y}.\frac{dy}{dx} = (-\sin x)(\log(\log x)) + \cos x.\frac{1}{\log x}.\frac{1}{x}\\ \frac{dy}{dx}= y( \cos x.\frac{1}{\log x}.\frac{1}{x}-\sin x\log(\log x) )\\ \frac{dy}{dx} = (\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )$
Therefore, the answer is $(\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )$

## Question:4 Differentiate the functions w.r.t. x. $x ^x - 2 ^{ \sin x }$

Given function is
$y = x ^x - 2 ^{ \sin x }$
Let's take $t = x^x$
take log on both the sides
$\log t=x\log x\\$
Now, differentiation w.r.t x is
$\log t=x\log x\\ \frac{d(\log t)}{dt}.\frac{dt}{dx} = \frac{d(x\log x)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)\\ \frac{1}{t}.\frac{dt}{dx} = \log x +1\\ \frac{dt}{dx} = t(\log x+1)\\ \frac{dt}{dx}= x^x(\log x+1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because t = x^x )$
Similarly, take $k = 2^{\sin x}$
Now, take log on both sides and differentiate w.r.t. x
$\log k=\sin x\log 2\\ \frac{d(\log k)}{dk}.\frac{dk}{dx} = \frac{d(\sin x\log 2)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)\\ \frac{1}{k}.\frac{dk}{dx} = \cos x \log 2\\ \frac{dk}{dx} = k(\cos x\log 2)\\ \frac{dk}{dx}= 2^{\sin x}(\cos x\log 2) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because k = 2^{\sin x} )$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}-\frac{dk}{dx}\\ \frac{dy}{dx} = x^x(\log x+1 )- 2^{\sin x}(\cos x\log 2)$

Therefore, the answer is $x^x(\log x+1 )- 2^{\sin x}(\cos x\log 2)$

## Question:5 Differentiate the functions w.r.t. x. $( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4$

Given function is
$y=( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4$
Take log on both sides
$\log y=\log [( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4]\\ \log y = 2\log(x+3)+3\log(x+4)+4\log(x+5)$
Now, differentiate w.r.t. x we get,
$\frac{1}{y}.\frac{dy}{dx} = 2.\frac{1}{x+3}+3.\frac{1}{x+4}+4.\frac{1}{x+5}\\ \frac{dy}{dx}=y\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )\\ \frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )\\ \frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right )\\ \frac{dy}{dx} = (x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)$
Therefore, the answer is $(x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)$

### Question:6 Differentiate the functions w.r.t. x. $( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }$

Given function is
$y = ( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }$
Let's take $t = ( x+ \frac{1}{x} ) ^ x$
Now, take log on both sides
$\log t =x \log ( x+ \frac{1}{x} )$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}.\frac{dt}{dx}=\log \left ( x+\frac{1}{x} \right )+x(1-\frac{1}{x^2}).\frac{1}{\left ( x+\frac{1}{x} \right )} = \frac{x^2-1}{x^2+1}+\log \left ( x+\frac{1}{x} \right )\\ \frac{dt}{dx} = t(\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))\\ \frac{dt}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))$
Similarly, take $k = x^{1+\frac{1}{x}}$
Now, take log on both sides
$\log k = ({1+\frac{1}{x}})\log x$
Now, differentiate w.r.t. x
We get,
$\frac{1}{k}.\frac{dk}{dx}=\frac{1}{x} \left ( 1+\frac{1}{x} \right )+(-\frac{1}{x^2}).\log x = \frac{x^2+1}{x^2}+\frac{-1}{x^2}.\log x\\ \frac{dk}{dx} = t(\frac{x^2+1}{x^2}+\left (\frac{-1}{x^2} \right )\log x)\\ \frac{dk}{dx} = x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$

Therefore, the answer is $\left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$

## Question:7 Differentiate the functions w.r.t. x. $(\log x )^x + x ^{\log x }$

Given function is
$y = (\log x )^x + x ^{\log x }$
Let's take $t = (\log x)^x$
Now, take log on both the sides
$\log t = x \log(\log x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} = \log (\log x) + x.\frac{1}{x}.\frac{1}{\log x}= \log (\log x)+\frac{1}{\log x}\\ \frac{dt}{dx}= t.(\log (\log x)+\frac{1}{\log x})\\ \frac{dt}{dx} =(\log x)^x(\log (\log x)) + (\log x)^x.\frac{1}{\log x}=(\log x)^x(\log (\log x))+ (\log x )^{x-1}$
Similarly, take $k = x^{\log x}$
Now, take log on both sides
$\log k = \log x \log x = (\log x)^2$
Now, differentiate w.r.t. x
We get,
$\frac{1}{k}\frac{dk}{dx} =2 (\log x).\frac{1}{x} \\ \frac{dt}{dx}= k.\left ( 2 (\log x).\frac{1}{x} \right )\\ \frac{dt}{dx} = x^{\log x}.\left (2 (\log x).\frac{1}{x} \right ) = 2x^{\log x-1}.\log x$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} =(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x$
Therefore, the answer is $(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x$

## Question:8 Differentiate the functions w.r.t. x. $(\sin x )^x + \sin ^{-1} \sqrt x$

Given function is
$(\sin x )^x + \sin ^{-1} \sqrt x$
Lets take $t = (\sin x)^x$
Now, take log on both the sides
$\log t = x \log(\sin x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} = \log (\sin x) + x.\cos x.\frac{1}{\sin x}= \log (\sin x)+x.\cot x \ \ \ (\because \frac{\cos x}{\sin x}=\cot x)\\ \frac{dt}{dx}= t.(\log (\sin x)+x.\cot x)\\ \frac{dt}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)$
Similarly, take $k = \sin^{-1}\sqrt x$
Now, differentiate w.r.t. x
We get,
$\frac{dk}{dt} = \frac{1}{\sqrt{1-(\sqrt x)^2}}.\frac{1}{2\sqrt x}= \frac{1}{2\sqrt{x-x^2}}\\ \frac{dk}{dt}=\frac{1}{2\sqrt{x-x^2}}\\$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}$
Therefore, the answer is $(\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}$

### Question:9 Differentiate the functions w.r.t. x $x ^ {\sin x } + ( \sin x )^{\cos x}$

Given function is
$y = x ^ { \sin x } + ( \sin x )^ {\cos x}$
Now, take $t = x^{\sin x}$
Now, take log on both sides
$\log t = \sin x \log x$
Now, differentiate it w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} = \cos x \log x+\frac{1}{x}.\sin x\\ \frac{dt}{dx}=t\left ( \cos x \log x+\frac{1}{x}.\sin x \right )\\ \frac{dt}{dx}= x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )$
Similarly, take $k = (\sin x)^{\cos x}$
Now, take log on both the sides
$\log k = \cos x \log (\sin x)$
Now, differentiate it w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dt} = (-\sin x)(\log (\sin x)) + \cos x.\frac{1}{\sin x}.\cos x=-\sin x\log(\sin x)+\cot x.\cos x\\ \frac{dk}{dt}= k\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )\\ \frac{dk}{dt}=(\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )$
Now,
$\frac{dy}{dx} = x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )+ (\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )$
Therefore, the answer is $x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )+ (\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )$

## Question:10 Differentiate the functions w.r.t. x. $x ^ {x \cos x} + \frac{x^2 + 1 }{x^2 -1 }$

Given function is
$x ^ {x \cos x} + \frac{x^2 + 1 }{x^2 -1 }$
Take $t = x^{x\cos x}$
Take log on both the sides
$\log t =x\cos x \log x$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} = \cos x\log x-x\sin x\log x + \frac{1}{x}.x.\cos x\\ \frac{dt}{dx}= t.\left (\log x(\cos x-x\sin x)+ \cos x \right ) = x^{x\cos x}\left ( \log x(\cos x-x\sin x)+ \cos x \right )$
Similarly,
take $k = \frac{x^2+1}{x^2-1}$
Now. differentiate it w.r.t. x
we get,
$\frac{dk}{dx} = \frac{2x(x^2-1)-2x(x^2+1)}{(x^2-1)^2} = \frac{2x^3-2x-2x^3-2x}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} = x^{x\cos x}\left ( \log x(\cos x-x\sin x)+ \cos x \right )-\frac{4x}{(x^2-1)^2}$
Therefore, the answer is $x^{x\cos x}\left ( \cos x(\log x+1)-x\sin x\log x\right )-\frac{4x}{(x^2-1)^2}$

### Question:11 Differentiate the functions w.r.t. x. $( x \cos x )^ x + ( x \sin x )^{1/ x}$

Given function is
$f(x)=( x \cos x )^ x + ( x \sin x )^{1/ x}$
Let's take $t = (x\cos x)^x$
Now, take log on both sides
$\log t =x\log (x\cos x) = x(\log x+\log \cos x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} =(\log x+\log \cos x)+x(\frac{1}{x}+\frac{1}{\cos x}.(-\sin x))\\ \frac{dt}{dx} = t(\log x + \log \cos x+1-x\tan x) \ \ \ \ \ \ \ \ \ (\because \frac{\sin x}{\cos x}= \tan x)\\ \frac{dt}{dx}= (x\cos x)^x(\log x + \log \cos x+1-x\tan x)\\ \frac{dt}{dx}=(x\cos x)^x(+1-x\tan x+\log (x\cos x))$
Similarly, take $k = (x\sin x)^{\frac{1}{x}}$
Now, take log on both the sides
$\log k = \frac{1}{x}(\log x+\log \sin x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dx} =(\frac{-1}{x^2})(\log x+\log \sin x)+\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.(\cos x))\\ \frac{dk}{dx} = \frac{k}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x}) \ \ \ \ \ \ \ \ \ (\because \frac{\cos x}{\sin x}= \cot x)\\ \frac{dk}{dx}=\frac{(x\sin x)^{\frac{1}{x}}}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})\\ \frac{dk}{dx}=(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$
Now,
$\frac{dy}{dx}= \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx}= (x\cos x)^x(+1-x\tan x+\log (x\cos x))+(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$
Therefore, the answer is $(x\cos x)^x(1-x\tan x+\log (x\cos x))+(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$

## Question:12 Find dy/dx of the functions given in Exercises 12 to 15

$x ^ y + y ^ x = 1$.

Given function is
$f(x)=x ^ y + y ^ x = 1$
Now, take $t = x^y$
take log on both sides
$\log t = y\log x$
Now, differentiate w.r.t x
we get,
$\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})\\ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})$
Similarly, take $k = y^x$
Now, take log on both sides
$\log k = x\log y$
Now, differentiate w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\ \frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})$
Now,
$f^{'}(x)= \frac{dt}{dx}+\frac{dk}{dx}= 0$

$( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) + (y^x)(\log y+\frac{x}{y}\frac{dy}{dx}) = 0\\ \frac{dy}{dx}(x^y(\log x)+xy^{x-1}) = -(yx^{y-1}+y^x(\log y))\\ \frac{dy}{dx}= \frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}$

Therefore, the answer is $\frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}$

## Question:13 Find dy/dx of the functions given in Exercises 12 to 15.

$y^x = x ^y$

Given function is
$f(x)\Rightarrow x ^ y = y ^ x$
Now, take $t = x^y$
take log on both sides
$\log t = y\log x$
Now, differentiate w.r.t x
we get,
$\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})\\ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})$
Similarly, take $k = y^x$
Now, take log on both sides
$\log k = x\log y$
Now, differentiate w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\ \frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})$
Now,
$f^{'}(x)\Rightarrow \frac{dt}{dx}= \frac{dk}{dx}$

$( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) = (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})\\ \frac{dy}{dx}(x^y(\log x)-xy^{x-1}) = (y^x(\log y)-yx^{y-1})\\ \frac{dy}{dx}= \frac{ y^x(\log y)-yx^{y-1}}{(x^y(\log x)-xy^{x-1})} = \frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )$

Therefore, the answer is $\frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )$

### Question:14 Find dy/dx of the functions given in Exercises 12 to 15. $( \cos x )^y = ( \cos y )^x$

Given function is
$f(x)\Rightarrow (\cos x) ^ y = (\cos y) ^ x$
Now, take log on both the sides
$y\log \cos x = x \log \cos y$
Now, differentiate w.r.t x
$\frac{dy}{dx}(\log \cos x)-y\tan x = \log \cos y-x\tan y\frac{dy}{dx}$
By taking similar terms on the same side
We get,
$(\frac{dy}{dx}(\log \cos x)-y\tan x) = (\log \cos y-x\tan y\frac{dy}{dx})\\ \frac{dy}{dx} \left (\log \cos x+(\cos y)^x.x\tan y) \right )= \left ( \log \cos y+(\cos x)^y.y\tan x \right )\\ \frac{dy}{dx}= \frac{\left (\log \cos y+y\tan x \right )}{\left ( \log \cos x+x\tan y) \right )} = \frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$

Therefore, the answer is $\frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$

### Question:15 Find dy/dx of the functions given in Exercises 12 to 15. $xy = e ^{x-y}$

Given function is
$f(x)\Rightarrow xy = e ^{x-y}$
Now, take log on both the sides
$\log x+\ log y = (x-y)(1) \ \ \ \ \ \ \ \ \ \ \ \ (\because \log e = 1)\\ \log x+\ log y = (x-y)$
Now, differentiate w.r.t x
$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$
By taking similar terms on same side
We get,
$(\frac{1}{y}+1)\frac{dy}{dx}=1-\frac{1}{x}\\ \frac{y+1}{y}.\frac{dy}{dx}= \frac{x-1}{x}\\ \frac{dy}{dx}= \frac{y}{x}.\frac{x-1}{y+1}$
Therefore, the answer is $\frac{y}{x}.\frac{x-1}{y+1}$

## Question:16 Find the derivative of the function given by $f (x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find

f ' (1)

Given function is
$y = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$
Take log on both sides
$\log y =\log (1 + x) + \log (1 + x^2) +\log (1 + x^4) +\log (1 + x^8)$
NOW, differentiate w.r.t. x
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8}\\ \frac{dy}{dx}=y.\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )\\ \frac{dy}{dx}= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )$
Therefore, $f^{'}(x)= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )$
Now, the vale of $f^{'}(1)$ is
$f^{'}(1)= (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8).\left ( \frac{1}{1+1}+ \frac{2(1)}{1+1^2}+ \frac{4(1)^3}{1+1^4}+ \frac{8(1)^7}{1+1^8} \right )\\ f^{'}(1)=16.\frac{15}{2} = 120$

## Question:17 (1) Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below: (i) by using product rule

Given function is
$f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Now, we need to differentiate using the product rule
$f^{'}(x)=\frac{d((x^2 - 5x + 8))}{dx}. (x^3 + 7x + 9)+(x^2 - 5x + 8).\frac{d( (x^3 + 7x + 9))}{dx}\\$
$= (2x-5).(x^3+7x+9)+(x^2-5x+8)(3x^2+7)\\ =2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56\\ = 5x^4 -20x^3+45x^2-52x+11$
Therefore, the answer is $5x^4 -20x^3+45x^2-52x+11$

### Question:17 (2) Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below: (ii) by expanding the product to obtain a single polynomial.

Given function is
$f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Multiply both to obtain a single higher degree polynomial
$f(x) = x^2(x^3+7x+9)-5x(x^3+7x+9)+8(x^3+7x+9)$
$= x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72$
$= x^5-5x^4+15x^3-26x^2+11x+72$
Now, differentiate w.r.t. x
we get,
$f^{'}(x)=5x^4-20x^3+45x^2-52x+11$
Therefore, the answer is $5x^4-20x^3+45x^2-52x+11$

### Question:17 (3) Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below: (iii) by logarithmic differentiation. Do they all give the same answer?

Given function is
$y=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Now, take log on both the sides
$\log y = \log (x^2-5x+8)+\log (x^3+7x+9)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{x^2-5x+8}.(2x-5) + \frac{1}{x^3+7x+9}.(3x^2+7)\\ \frac{dy}{dx}= y.\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\ \frac{dy}{dx}=(x^2-5x+8)(x^3+7x+9).\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\ \frac{dy}{dx} = (2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)\\ \frac{dy}{dx} = 5x^4-20x^3+45x^2-56x+11$
Therefore, the answer is $5x^4-20x^3+45x^2-56x+11$
And yes they all give the same answer

## Question:18 If u, v and w are functions of x, then show that $\frac{d}{dx} ( u,v,w) = \frac{du}{dx} v. w +u . \frac{dv }{dx } v. w+ u . \frac{dv}{dx } . w+u.v \frac{dw}{dx}$ in two ways - first by repeated application of product rule, second by logarithmic differentiation.

It is given that u, v and w are the functions of x
Let $y = u.v.w$
Now, we differentiate using product rule w.r.t x
First, take $y = u.(vw)$
Now,
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{d(v.w)}{dx}.u$ -(i)
Now, again by the product rule
$\frac{d(v.w)}{dx}= \frac{dv}{dx}.w + \frac{dw}{dx}.v$
Put this in equation (i)
we get,
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)$
Hence, by product rule we proved it

Now, by taking the log
Again take $y = u.v.w$
Now, take log on both sides
$\log y = \log u + \log v + \log w$
Now, differentiate w.r.t. x
we get,
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{u}.\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\\ \frac{dy}{dx}= y. \left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\ \frac{dy}{dx} = (u.v.w)\left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\$
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)$
Hence, we proved it by taking the log

## More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5:-

Class 12 Maths ch 5 ex 5.5 consists of questions related to finding differentiation of functions raised to the power of functions. These types of questions can be solved using the concept called logarithmic differentiation. In NCERT book exercise 5.5 Class 12 Maths which you learn this concept through solving problems based on this concept. There are 10 questions from this concept given in the exercise 5.5 class 12 maths. Also, you can solve two examples give before this exercise which will help you to get conceptual clarity. The proof of this concept is also given before this exercise. You can prove the given definition by yourself using the chain rule and logarithmic property.

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

## Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5:-

• NCERT Solutions for Class 12 Maths chapter 5 exercise 5.5 are beneficial when students are getting difficulty while solving the NCERT syllabus exercise 5.5 Class 12 Maths problems.
• Class 12 Maths chapter 5 exercise 5.5 solutions are descriptive so you can easily understand the solutions.
• Class 12th Maths chapter 5 exercise 5.5 solutions can be used for reference when you are solving logarithmic differentiation.

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

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