RD Sharma Class 12 Ex 11.6 Solutions Chapter 11 Differentiation

Here we provide RD Sharma Class 12 Ex 11.6 Solutions Chapter 11 Differentiation for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 11.6 Solutions Chapter 11 Differentiation book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter11
Exercise11.6
CategoryRD Sharma Solutions

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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