Class 12 RD Sharma Solutions Chapter 11 Differentiation Exercise 11.6 (original) (raw)

Last Updated : 23 Jul, 2025

Chapter 11 of RD Sharma's Class 12 Mathematics textbook focuses on the Differentiation a fundamental concept in calculus. Differentiation is the process of finding the rate at which a function changes at any point. It is essential for understanding the behavior of the functions particularly in fields like physics, engineering, and economics. Exercise 11.6 in this chapter specifically deals with applying differentiation rules to solve various problems.

Differentiation

The Differentiation is the mathematical process of finding the derivative of a function. The derivative represents the rate of change of the function with respect to a variable. It is a key concept in calculus helping in the analyzing and understanding the behavior of functions such as finding slopes of curves, and velocities and optimizing functions.

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)}

**Read More:

• Differentiation
• Derivative Rules

Summary

Exercise 11.6 in RD Sharma's Class 12 Mathematics textbook focuses on the application of derivatives to find the errors and approximations in measurements. This set covers problems involving the concept of relative error, percentage error, and approximations using differentials. Students learn to apply the differentiation techniques to estimate the change in a function's value due to small changes in its variables, which is crucial in practical applications of calculus in physics, engineering, and other sciences