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. $\cos x . \cos 2x .\cos 3x$

Answer:

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.

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

Answer:

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. $(\log x ) ^{\cos x}$

Answer:

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. $x ^x - 2 ^{ \sin x }$

Answer:

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. $( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4$

Answer:

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. $( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }$

Answer:

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. $(\log x )^x + x ^{\log x }$

Answer:

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. $(\sin x )^x + \sin ^{-1} \sqrt x$

Answer:

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 $x ^ {\sin x } + ( \sin x )^{\cos x}$

Answer:

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. $x ^ {x \cos x} + \frac{x^2 + 1 }{x^2 -1 }$

Answer:

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. $( x \cos x )^ x + ( x \sin x )^{1/ x}$

Answer:

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

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

Answer:

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.

$y^x = x ^y$

Answer:

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. $( \cos x )^y = ( \cos y )^x$

Answer:

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. $xy = e ^{x-y}$

Answer:

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)

Answer:

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

Answer:

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.

Answer:

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?

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.

Answer:

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

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

Happy learning!!!

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

Continuity and Differentiability Exercise: 5.5

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

Question:8 Differentiate the functions w.r.t. x.

Answer:

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

Answer:

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

Answer:

Question:11 Differentiate the functions w.r.t. x.

Answer:

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

Answer:

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

Answer:

Question:14 Find dy/dx of the functions given in Exercises 12 to 15.

Answer:

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

Answer:

Question:16 Find the derivative of the function given by and hence find

Answer:

Question:17 (1) Differentiate in three ways mentioned below: (i) by using product rule

Answer:

Question:17 (2) Differentiate in three ways mentioned below: (ii) by expanding the product to obtain a single polynomial.

Answer:

Question:17 (3) Differentiate in three ways mentioned below: (iii) by logarithmic differentiation. Do they all give the same answer?

Answer:

Question:18 If u, v and w are functions of x, then show that in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Answer:

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

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

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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?

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.