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

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

$\log y = \frac{1}{2} \log\left( \frac{(x - 1)(x - 2)}{(x - 3)(x - 4)(x - 5)} \right)$

$\log y = \frac{1}{2} \left( \log(x - 1) + \log(x - 2) - \log(x - 3) - \log(x - 4) - \log(x - 5) \right)$
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})$

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{2} \left( \frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5} \right)$

$\frac{dy}{dx} = y \cdot \frac{1}{2} \left( \frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5} \right)$

$\frac{dy}{dx} = \frac{1}{2} \sqrt{ \frac{(x - 1)(x - 2)}{(x - 3)(x - 4)(x - 5)} } \left( \frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5} \right)$

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

$\frac{d(\log y)}{dx} = \frac{d(\cos x \log(\log x))}{dx}$

$\frac{1}{y} \cdot \frac{dy}{dx} = (-\sin x)(\log(\log x)) + \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x}$

$\frac{dy}{dx} = y \left( \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x} - \sin x \log(\log x) \right)$

$\frac{dy}{dx} = (\log x)^{\cos x} \left( \frac{\cos x}{x \log x} - \sin x \log(\log x) \right)$

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

$\log t = x \log x$

$\frac{d(\log t)}{dt} \cdot \frac{dt}{dx} = \frac{d(x \log x)}{dx} \ \ \ \ \ \ \ (\text{by chain rule})$

$\frac{1}{t} \cdot \frac{dt}{dx} = \log x + 1$

$\frac{dt}{dx} = t(\log x + 1)$

$\frac{dt}{dx} = x^x(\log x + 1) \ \ \ \ \ \ \ \ \ \ \ \ \ (\because t = x^x)$

Similarly, take $k = 2^{\sin x}$
Now, take log on both sides and differentiate w.r.t. x

$\log k = \sin x \log 2$

$\frac{d(\log k)}{dk} \cdot \frac{dk}{dx} = \frac{d(\sin x \log 2)}{dx} \ \ \ \ \ \ \ (\text{by chain rule})$

$\frac{1}{k} \cdot \frac{dk}{dx} = \cos x \log 2$

$\frac{dk}{dx} = k(\cos x \log 2)$

$\frac{dk}{dx} = 2^{\sin x}(\cos x \log 2) \ \ \ \ \ \ \ \ \ \ \ \ \ (\because k = 2^{\sin x})$

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,

$\frac{1}{y} \cdot \frac{dy}{dx} = 2 \cdot \frac{1}{x+3} + 3 \cdot \frac{1}{x+4} + 4 \cdot \frac{1}{x+5}$

$\frac{dy}{dx} = y\left( \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right)$

$\frac{dy}{dx} = (x+3)^2 (x+4)^3 (x+5)^4 \left( \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right)$

$\frac{dy}{dx} = (x+3)^2 (x+4)^3 (x+5)^4 \left( \frac{2(x+4)(x+5) + 3(x+3)(x+5) + 4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right)$

$\frac{dy}{dx} = (x+3)(x+4)^2(x+5)^3(9x^2 + 70x + 133)$

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,

$\frac{1}{t} \cdot \frac{dt}{dx} = \log \left( x + \frac{1}{x} \right) + x\left(1 - \frac{1}{x^2} \right) \cdot \frac{1}{\left( x + \frac{1}{x} \right)}$

$= \frac{x^2 - 1}{x^2 + 1} + \log \left( x + \frac{1}{x} \right)$

$\frac{dt}{dx} = t \left( \frac{x^2 - 1}{x^2 + 1} + \log \left( x + \frac{1}{x} \right) \right)$

$\frac{dt}{dx} = \left( x + \frac{1}{x} \right)^x \left( \frac{x^2 - 1}{x^2 + 1} + \log \left( x + \frac{1}{x} \right) \right)$

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,

$\frac{1}{k} \cdot \frac{dk}{dx} = \frac{1}{x} \left( 1 + \frac{1}{x} \right) + \left(-\frac{1}{x^2} \right) \log x$

$= \frac{x^2 + 1}{x^2} + \left( -\frac{1}{x^2} \right) \log x$

$\frac{dk}{dx} = k \left( \frac{x^2 + 1 - \log x}{x^2} \right)$

$\frac{dk}{dx} = x^{x + \frac{1}{x}} \left( \frac{x^2 + 1 - \log x}{x^2} \right)$

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,

$\frac{1}{t} \cdot \frac{dt}{dx} = \log (\log x) + x \cdot \frac{1}{x} \cdot \frac{1}{\log x} = \log (\log x) + \frac{1}{\log x}$

$\frac{dt}{dx} = t \cdot \left( \log (\log x) + \frac{1}{\log x} \right)$

$\frac{dt}{dx} = (\log x)^x \cdot \log (\log x) + (\log x)^x \cdot \frac{1}{\log x}$

$\frac{dt}{dx} = (\log x)^x \cdot \log (\log x) + (\log x)^{x-1}$

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,

$\frac{1}{t} \cdot \frac{dt}{dx} = \log (\sin x) + x \cdot \cos x \cdot \frac{1}{\sin x} = \log (\sin x) + x \cdot \cot x \ \ \ (\because \frac{\cos x}{\sin x} = \cot x)$

$\frac{dt}{dx} = t \cdot (\log (\sin x) + x \cdot \cot x)$

$\frac{dt}{dx} = (\sin x)^x \cdot (\log (\sin x) + x \cdot \cot x)$

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,

$\frac{1}{t} \cdot \frac{dt}{dx} = \cos x \cdot \log x + \frac{1}{x} \cdot \sin x$

$\frac{dt}{dx} = t \left( \cos x \cdot \log x + \frac{1}{x} \cdot \sin x \right)$

$\frac{dt}{dx} = x^{\sin x} \left( \cos x \cdot \log x + \frac{1}{x} \cdot \sin x \right)$

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,
$\frac{1}{k}\frac{dk}{dx} = (-\sin x)(\log (\sin x)) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x = -\sin x \log(\sin x) + \cot x \cdot \cos x$
$\frac{dk}{dx} = k\left( -\sin x \log(\sin x) + \cot x \cdot \cos x \right)$
$\frac{dk}{dx} = (\sin x)^{\cos x}\left( -\sin x \log(\sin x) + \cot x \cdot \cos x \right)$
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,

$\frac{1}{t} \cdot \frac{dt}{dx} = \cos x \cdot \log x - x \cdot \sin x \cdot \log x + \frac{1}{x} \cdot x \cdot \cos x$

$\frac{dt}{dx} = t \cdot \left( \log x (\cos x - x \sin x) + \cos x \right)$

$\frac{dt}{dx} = x^{x \cos x} \cdot \left( \log x (\cos x - x \sin x) + \cos x \right)$

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,

$\frac{1}{t} \cdot \frac{dt}{dx} = (\log x + \log \cos x) + x \left( \frac{1}{x} + \frac{1}{\cos x} \cdot (-\sin x) \right)$

$\frac{dt}{dx} = t \left( \log x + \log \cos x + 1 - x \tan x \right) \ \ \ \ \ \ (\because \frac{\sin x}{\cos x} = \tan x)$

$\frac{dt}{dx} = (x \cos x)^x \left( \log x + \log \cos x + 1 - x \tan x \right)$

$\frac{dt}{dx} = (x \cos x)^x \left( 1 - x \tan x + \log(x \cos x) \right)$

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,
$\frac{1}{k}\frac{dk}{dx} = \left(\frac{-1}{x^2}\right)(\log x + \log \sin x) + \frac{1}{x} \left(\frac{1}{x} + \frac{1}{\sin x} \cdot \cos x\right)$
$\frac{dk}{dx} = \frac{k}{x^2} \left(-\log x - \log \sin x + \frac{1}{x^2} + \frac{\cot x}{x} \right) \ \ \ \ \ \ \ \ \ (\because \frac{\cos x}{\sin x} = \cot x)$
$\frac{dk}{dx} = \frac{(x \sin x)^{\frac{1}{x}}}{x^2} \left(-\log x - \log \sin x + \frac{1}{x^2} + \frac{\cot x}{x} \right)$
$\frac{dk}{dx} = (x \sin x)^{\frac{1}{x}} \cdot \frac{x \cot x + 1 - \log(x \sin x)}{x^2}$
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,

$\frac{1}{t} \cdot \frac{dt}{dx} = \frac{dy}{dx} \cdot (\log x) + y \cdot \frac{1}{x} = \frac{dy}{dx} \cdot (\log x) + \frac{y}{x}$

$\frac{dt}{dx} = t \left( \frac{dy}{dx} \cdot (\log x) + \frac{y}{x} \right)$

$\frac{dt}{dx} = x^y \left( \frac{dy}{dx} \cdot (\log x) + \frac{y}{x} \right)$

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)\left( \frac{dy}{dx} \log x + \frac{y}{x} \right) + (y^x)\left( \log y + \frac{x}{y} \frac{dy}{dx} \right) = 0 $

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

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

$\frac{1}{t} \cdot \frac{dt}{dx} = \frac{dy}{dx} \cdot \log x + y \cdot \frac{1}{x} = \frac{dy}{dx} \cdot \log x + \frac{y}{x}$

$\frac{dt}{dx} = t \left( \frac{dy}{dx} \cdot \log x + \frac{y}{x} \right)$

$\frac{dt}{dx} = x^y \left( \frac{dy}{dx} \cdot \log x + \frac{y}{x} \right)$

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,
$f^{'}(x)\Rightarrow \frac{dt}{dx}= \frac{dk}{dx}$

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

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,
$\left( \frac{dy}{dx} (\log \cos x) - y \tan x \right) = \left( \log \cos y - x \tan y \frac{dy}{dx} \right)$
$\frac{dy}{dx} \left( \log \cos x + x \tan y \right) = \log \cos y + y \tan x$
$\frac{dy}{dx} = \frac{y \tan x + \log \cos y}{x \tan y + \log \cos x}$

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

$\log x + \log y = (x - y)(1) \qquad (\because \log e = 1)$

$\log x + \log y = x - y$

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

$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8}$

$\frac{dy}{dx} = y \cdot \left( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right)$

$\frac{dy}{dx} = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8) \cdot \left( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right)$

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 value 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,

$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x^2 - 5x + 8} \cdot (2x - 5) + \frac{1}{x^3 + 7x + 9} \cdot (3x^2 + 7)$

$\frac{dy}{dx} = y \cdot \left( \frac{(2x - 5)(x^3 + 7x + 9) + (3x^2 + 7)(x^2 - 5x + 8)}{(x^2 - 5x + 8)(x^3 + 7x + 9)} \right)$

$\frac{dy}{dx} = (x^2 - 5x + 8)(x^3 + 7x + 9) \cdot \left( \frac{(2x - 5)(x^3 + 7x + 9) + (3x^2 + 7)(x^2 - 5x + 8)}{(x^2 - 5x + 8)(x^3 + 7x + 9)} \right)$

$\frac{dy}{dx} = (2x - 5)(x^3 + 7x + 9) + (3x^2 + 7)(x^2 - 5x + 8)$

$\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 56x + 11$

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,

$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} + \frac{1}{w} \cdot \frac{dw}{dx}$

$\frac{dy}{dx} = y \cdot \left( \frac{v w \cdot \frac{du}{dx} + u w \cdot \frac{dv}{dx} + u v \cdot \frac{dw}{dx}}{u v w} \right)$

$\frac{dy}{dx} = (u v w) \cdot \left( \frac{v w \cdot \frac{du}{dx} + u w \cdot \frac{dv}{dx} + u v \cdot \frac{dw}{dx}}{u v w} \right)$

$\frac{dy}{dx} = \frac{du}{dx} \cdot (v w) + \frac{dv}{dx} \cdot (u w) + \frac{dw}{dx} \cdot (u v)$

Hence, we proved it by taking the log