RD Sharma Class 12 Ex 11.6 Solutions Chapter 11 Differentiation

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	11
Exercise	11.6
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 11.6 Solutions Chapter 11 Differentiation

Question 1. If $y=\sqrt{x+\sqrt{x+\sqrt{x+....to\ \infin}}}$ , prove that $\frac{dy}{dx}=\frac{1}{2y-1}$

Solution:

We have, $y=\sqrt{x+\sqrt{x+\sqrt{x+....to\ \infin}}}$

⇒ $y=\sqrt{x+y}$

Squaring both sides, we get,

y² = x + y

$⇒2y\frac{dy}{dx}=1+\frac{dy}{dx}\\ ⇒\frac{dy}{dx}(2y-1)=1\\ ⇒\frac{dy}{dx}=\frac{1}{2y-1}$

Question 2. If $y=\sqrt{cosx+\sqrt{cosx+\sqrt{cosx+ ....\ to\ \infin}}}$ , prove that $\frac{dy}{dx}=\frac{sinx}{1-2y}$

Solution:

We have, $y=\sqrt{cosx+\sqrt{cosx+\sqrt{cosx+ ....\ to\ \infin}}}$
⇒ $y = \sqrt{cosx+y}$
Squaring both sides, we get,
y² = cos x + y
⇒ $2y\frac{dy}{dx}=-sinx+\frac{dy}{dx}\\ ⇒ \frac{dy}{dx}(2y-1)=-sinx\\ ⇒ \frac{dy}{dx}=\frac{-sinx}{1-2y}\\ ⇒ \frac{dy}{dx}=\frac{sinx}{1-2y}$

Question 3. If $y=\sqrt{logx+\sqrt{logx+\sqrt{logx+ ....\ to\ \infin}}}$ , prove that $(2y-1)\frac{dy}{dx}=\frac{1}{x}$

Solution:

We have, $y=\sqrt{logx+\sqrt{logx+\sqrt{logx+ ....\ to\ \infin}}}$

⇒ $y=\sqrt{logx+y}$

Squaring both sides, we get,

y² = log x + y

$⇒ 2y\frac{dy}{dx}=\frac{1}{x}+\frac{dy}{dx}\\ ⇒ \frac{dy}{dx}(2y-1)=\frac{1}{x}$

Question 4. If $y=\sqrt{tanx+\sqrt{tanx+\sqrt{tanx+ ....\ to\ \infin}}}$ , prove that $\frac{dy}{dx}=\frac{sec^2x}{2y-1}$

Solution:

We have, $y=\sqrt{tanx+\sqrt{tanx+\sqrt{tanx+ ....\ to\ \infin}}}$

⇒ $y =\sqrt{tanx+y}$

Squaring both sides, we get,

y² = tan x + y

$⇒2y\frac{dy}{dx}=sec^2x+\frac{dy}{dx}\\ ⇒\frac{dy}{dx}(2y-1)=sec^2x\\ ⇒\frac{dy}{dx}=\frac{sec^2x}{2y-1}$

Question 5. If $y=(sinx)^{(sinx)^{(sinx)^{...\infin}}}$ , prove that $\frac{dy}{dx}=\frac{y^2cotx}{(1-y\ logsinx)}$

Solution:

We have, $y=(sinx)^{(sinx)^{(sinx)^{...\infin}}}$

⇒ y = (sin x)^y

Taking log on both sides,

log y = log(sin x)^y

⇒ log y = y log(sin x)

$⇒\frac{1}{y}\frac{dy}{dx}=y\frac{d}{dx}\{log(sinx)\}+log\ sinx\frac{dy}{dx}\\ ⇒\frac{1}{y}\frac{dy}{dx}=y\left(\frac{1}{sinx}\right)\frac{d}{dx}(sinx)+log\ sinx\frac{dy}{dx}\\ ⇒\frac{dy}{dx}\left(\frac{1}{y}-log\ sinx\right)=\frac{y}{sinx}(cosx)\\ ⇒\frac{dy}{dx}\left(\frac{1-y\ log\ sinx}{y}\right)=y\ cotx\\ ⇒\frac{dy}{dx}=\frac{y^2cotx}{1-y\ log\ sinx}$

Question 6. If $y=(tanx)^{(tanx)^{(tanx)^{...\infin}}}$ , prove that $\frac{dy}{dx}=2\ at\ x=\frac{\pi}{4}$

Solution:

We have, $y=(tanx)^{(tanx)^{(tanx)^{...\infin}}}$

⇒ y = (tan x)^y

Taking log on both sides,

log y = log(tan x)^y

⇒ log y = y log tan x

Differentiating with respect to x using chain rule,

$⇒ \frac{1}{y}\frac{dy}{dx}=y\frac{d}{dx}\{\log\ tanx\}+\log\ tan\frac{dy}{dx}\\ ⇒ \frac{1}{y}\frac{dy}{dx}=\frac{y}{tanx}\frac{d}{dx}\{tanx\}+\log\ tan\frac{dy}{dx}\\ ⇒\frac{dy}{dx}\left(\frac{1}{y}-\log tanx\right)=\frac{y}{tanx}sec^2x\\ ⇒\frac{dy}{dx}=\frac{y}{tanx}sec^2x\times\left(\frac{y}{1-y\log tanx}\right)\\$

Now,

$\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}}=\frac{y\ sec^2\left(\frac{\pi}{4}\right)}{tan\left(\frac{\pi}{4}\right)}\times\frac{y}{1-y\ \log tan\left(\frac{\pi}{4}\right)}\\ ⇒\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}}=\frac{y^2(\sqrt{2})^2}{1(1-y\log\ tan1)}\\ ⇒\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}}=\frac{2(1)^2}{(1-0)}\ \ \ \ \ \left[\because\ (y)_{\frac{\pi}{4}}=\left(tan\frac{\pi}{4}\right)^{\left(tan\frac{\pi}{4}\right)^{\left(tan\frac{\pi}{4}\right)^{...\infin}}}=1\right]\\ ⇒\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}}=2$

Question 7. If $y=e^{x^{e^x}}+x^{e^{x^e}}+e^{x^{x^e}}$ , prove that $e^{x^{e^x}}\times x^{e^{x}}\left\{\frac{e^x}{x}+e^x\times \log x\right\}+e^{e^{e^x}}\times e^{e^{^x}}\left\{\frac{1}{x}+e^x\times \log x\right\}+e^{x^{x^e}}x^{x^{e}}\times x^{e-1}\{1+e\log x\}$

Solution:

We have, $y=e^{x^{e^x}}+x^{e^{x^e}}+e^{x^{x^e}}$

⇒ y = u + v + w

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}\ \ ...(i)$

where $u=e^{x^{e^x}},\ v=x^{e^{x^e}}\ and\ w=e^{x^{x^e}}$

Now, $u=e^{x^{e^x}}\ \ \ \ \ ....(ii)$

Taking log on both sides,

$\log u=\log e^{x^{e^x}}\\ \Rightarrow\log u=x^{e^{x}}\log e\\ \Rightarrow\log u=x^{e^{x}}\ \ \ ...(iii)\\ \log \log u=\log x^{e^{x}}\\ \Rightarrow\log \log u = e^x\log x$

Differentiating with respect to x,

$\Rightarrow\frac{1}{\log u}\frac{d}{dx}(\log u)=e^x\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(e^x)\\ \Rightarrow\frac{1}{\log u}\frac{1}{u}\frac{du}{dx}=\frac{e^x}{x}+e^x\log x\\ \Rightarrow\frac{du}{dx}=u\log u\left[\frac{e^x}{x}+e^x\log x\right]\\ \Rightarrow\frac{du}{dx}=e^{x^{e^x}}\times x^{e^{x}}\left[\frac{e^x}{x}+e^x\log x\right]\\ Now,\ v=x^{e^{x^e}}\ \ \ \ \ ...(iv)$

Taking log on both sides,

$\log v=\log x^{e^{e^x}}\\ \Rightarrow\log v = e^{e^{x}}\log x$

Taking log on both sides

$\Rightarrow\frac{1}{\log w}\frac{d}{dx}(\log w)=x^e\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x^e)\\ \Rightarrow\frac{1}{\log w}\left(\frac{1}{w}\right)\frac{dw}{dx}=x^e\left(\frac{1}{x}\right)+\log xex^{e-1}\\ \Rightarrow\frac{dw}{dx}=w\log w[x^{e-1}+e\log xx^{e-1}]\\ \Rightarrow\frac{dw}{dx}=e^{x^{x^e}}x^{x^{e}}x^{e-1}(1+e\log x)\\$

Using equation in equation (i), we get

$\frac{dy}{dx}=e^{x^{e^x}}{x^{e^x}}\left[\frac{e^x}{x}+e^x\log x\right]+x^{e^{e^x}}\times e^{e^x}\left[\frac{1}{x}+e^x\log x\right]+e^{x^{x^e}}{x^{x^e}}x^{e-1}(1+e\log x)$

Question 8. If $y=(cosx)^{(cosx)^{(cosx)^\ ...\infin}}$ , Prove that $\frac{dy}{dx}=\frac{y^2tanx}{(1-y\log cosx)}$

Solution:

We have, $y=(cosx)^{(cosx)^{(cosx)^\ ...\infin}}$

⇒ y = (cos x)^y

Taking log on both sides,

log y = log(cos x)^y

⇒ log y = y log (cos x)

Differentiating with respect to x using chain rule,

$\frac{1}{y}\frac{dy}{dx}=y\frac{d}{dx}\{\log cosx\}+\log cos x\frac{dy}{dx}\\ \Rightarrow\frac{1}{y}\frac{dy}{dx}=y\left(\frac{1}{cosx}\right)\frac{d}{dx}(cosx)+\log cosx\frac{dy}{dx}\\ \Rightarrow\frac{dy}{dx}\left(\frac{1}{y}-\log cosx\right)=\frac{y}{cosx}(-sinx)\\ \Rightarrow\frac{dy}{dx}\left(\frac{1-y\log cosx}{y}\right)=-y\ tanx\\ \Rightarrow\frac{dy}{dx}=-\frac{y^2tanx}{(1-y\log cosx)}$

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RD Sharma Class 12 Ex 11.6 Solutions Chapter 11 Differentiation

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Question 2. If , prove that

Question 3. If , prove that

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Question 6. If , prove that

Question 7. If , prove that

Question 8. If , Prove that

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Question 1. If $y=\sqrt{x+\sqrt{x+\sqrt{x+....to\ \infin}}}$ , prove that $\frac{dy}{dx}=\frac{1}{2y-1}$

Question 2. If $y=\sqrt{cosx+\sqrt{cosx+\sqrt{cosx+ ....\ to\ \infin}}}$ , prove that $\frac{dy}{dx}=\frac{sinx}{1-2y}$

Question 3. If $y=\sqrt{logx+\sqrt{logx+\sqrt{logx+ ....\ to\ \infin}}}$ , prove that $(2y-1)\frac{dy}{dx}=\frac{1}{x}$

Question 4. If $y=\sqrt{tanx+\sqrt{tanx+\sqrt{tanx+ ....\ to\ \infin}}}$ , prove that $\frac{dy}{dx}=\frac{sec^2x}{2y-1}$

Question 5. If $y=(sinx)^{(sinx)^{(sinx)^{...\infin}}}$ , prove that $\frac{dy}{dx}=\frac{y^2cotx}{(1-y\ logsinx)}$

Question 6. If $y=(tanx)^{(tanx)^{(tanx)^{...\infin}}}$ , prove that $\frac{dy}{dx}=2\ at\ x=\frac{\pi}{4}$

Question 7. If $y=e^{x^{e^x}}+x^{e^{x^e}}+e^{x^{x^e}}$ , prove that $e^{x^{e^x}}\times x^{e^{x}}\left\{\frac{e^x}{x}+e^x\times \log x\right\}+e^{e^{e^x}}\times e^{e^{^x}}\left\{\frac{1}{x}+e^x\times \log x\right\}+e^{x^{x^e}}x^{x^{e}}\times x^{e-1}\{1+e\log x\}$

Question 8. If $y=(cosx)^{(cosx)^{(cosx)^\ ...\infin}}$ , Prove that $\frac{dy}{dx}=\frac{y^2tanx}{(1-y\log cosx)}$