What makes RD Sharma books better than NCERT?

RD Sharma books are generally more detailed and exam-oriented when it comes to maths. NCERT materials are also suitable, but they are only meant for a basic understanding of the subject.

Will I be able to solve NCERT differentiation questions after solving this material?

The general concept of these chapters remains the same no matter which material students choose. Therefore, once they are done preparing from RD Sharma Class 12 solutions Differentiation Ex 10.6 material, they can quickly solve NCERT questions.

What is a derivative?

The change in the rate of a function is called a derivative. To learn more about differentiation, check out RD Sharma Class 12 Chapter 10 Exercise 10.6.

What is the opposite of differentiation?

Integration is the opposite of differentiation. The same value is obtained after integrating a derivative. To know more about this chapter, check RD Sharma Class 12th Exercise 10.6.

Can I score better marks with this material?

Practicing this material regularly will help students cover more of the syllabus and study efficiently. This will increase their understanding and, in turn, help them score better marks.

RD Sharma Class 12 Exercise 10.6 Differentiation Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 10.6 Differentiation Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 10.6 Differentiation Solutions Maths - Download PDF Free Online

Updated on 20 Jan 2022, 05:32 PM IST

RD Sharma materials are well known among Indian schools as they are one of the best materials for maths. These materials also surpass NCERT in terms of chapter details and exam-oriented questions.
RD Sharma Class 12th Exercise 10.6 contains the concepts for Differentiation. This is a relatively small exercise containing eight questions only. These include Level 1 questions that have proof sums and evaluation. As the sums deal with the same theorems, students can solve a few problems for their understanding and then refer to RD Sharma solutions material as it is more convenient.

This Story also Contains

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise
Differentiation Excercise: 10.6
RD Sharma Chapter-wise Solutions

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise

Differentiation Excercise: 10.6

Differentiation exercise 10.6 question 1

Answer: $\frac{d y}{d x}=\frac{1}{2 y-1}$
Hint: The value of y is given as infinite series. If a term is deleted from an infinite series, it remains the same in this case.
Given: $y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots+\infty}}}$
Solution:
Here it is given that,
$y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots+\infty}}}$
This can be written as:
$y=\sqrt{x+y}$
Squaring on both sides, we get:
$y^{2}=x+y$ …(1)
Differentiating (1) w.r.t x,
$\begin{aligned} &2 y \frac{d y}{d x}=1+\frac{d y}{d x} \\ &\frac{d y}{d x}(2 y-1)=1 \\ &\therefore \frac{d y}{d x}=\frac{1}{2 y-1} \end{aligned}$
Hence, it is proved.

Differentiation exercise 10.6 question 2

Answer: $\frac{d y}{d x}=\frac{\sin x}{1-2 y}$
Hint: The value of y is given as infinite series. If a term is deleted from an infinite series, it remains the same in this case.
Given: $y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \ldots \ldots . .+\infty}}}$
Solution:
Here it is given that,
$y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \ldots \ldots . .+\infty}}}$
This can be written as:
$y=\sqrt{\cos x+y}$
Squaring on both sides, we get:
$y^{2}=\left ( \cos x+y \right )$ …(1)
Differentiating (1) w.r.t x,
$\begin{aligned} &2 y \frac{d y}{d x}=-\sin x+\frac{d y}{d x} \\ &\frac{d y}{d x}(2 y-1)=-\sin x \\ &\frac{d y}{d x}=\frac{-\sin x}{2 y-1} \end{aligned}$
$\therefore \frac{d y}{d x}=\frac{-\sin x}{2 y-1}$
Hence, it is proved.

Differentiation exercise 10.6 question 3

Answer: $\left ( 2y-1 \right )\frac{dy}{dx}=\frac{1}{x}$
Hint: The value of y is given as infinite series. If a term is deleted from an infinite series, it remains the same in this case.
Given: $y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \ldots \ldots .+\infty}}}$
Solution:
Here it is given that,
$y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \ldots \ldots .+\infty}}}$
This can be written as:
$y=\sqrt{\log x+y}$
Squaring on both sides, we get:
$y^{2}=\left ( \log x+y \right )$ …(1)
Differentiating (1) w.r.t x,
$\begin{aligned} &2 y \frac{d y}{d x}=\frac{1}{x}+\frac{d y}{d x} \\ &\frac{d y}{d x}(2 y-1)=\frac{1}{x} \end{aligned}$
$\therefore \frac{d y}{d x}(2 y-1)=\frac{1}{x}$
. Hence, it is proved.

Differentiation exercise 10.6 question 4

Answer: $\frac{dy}{dx}=\frac{\sec ^{2}x}{2y-1}$
Hint: The value of y is given as infinite series. If a term is deleted from an infinite series, it remains the same in this case.
Given: $y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \ldots \ldots .+\infty}}}$
Solution:
Here it is given that,
$y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \ldots \ldots .+\infty}}}$
This can be written as:
$y=\sqrt{\tan x+y}$
Squaring on both sides, we get:
$y^{2}=\left ( \tan x+y \right )$ …(1)
Differentiating (1) w.r.t x,
$2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x}$
$\frac{d y}{d x}\left ( 2y-1 \right )=\sec ^{2} x$
$\therefore \frac{d y}{d x}=\frac{\sec ^{2} x}{ 2y-1}$
Hence, it is proved.

Differentiation exercise 10.6 question 5

Answer: $\frac{d y}{d x}=\frac{y^{2} \cot x}{1-y \log \sin x}$
Hint: The value of y is given as infinite series. If a term is deleted from an infinite series, it remains the same in this case.
Given: $y=\left ( \sin x \right )^{\left ( \sin \right )^{..... \infty }}$
Solution:
Here it is given that,
$y=\left ( \sin x \right )^{\left ( \sin \right )^{..... \infty }}$
This can be written as:
$\log y=\log \left ( \sin x \right )^{y}$
Taking log on both sides, we get:
$\log y=\log \left ( \sin x \right )^{y}$
$\therefore \log y=y\log \left ( \sin x \right )$ …(1)
Differentiating (1) w.r.t x,
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=y \frac{d}{d x}(\log \sin x)+\log \sin x \cdot \frac{d}{d x}(y) \\ &\frac{1}{y} \frac{d y}{d x}=y \frac{1}{\sin x}(\cos x)+\log \sin x \frac{d y}{d x} \\ &\frac{d y}{d x}\left(\frac{1}{y}-\log \sin x\right)=y \cot x \\ &\frac{d y}{d x}\left(\frac{1-y \log \sin x}{y}\right)=y \cot x \\ &\therefore \frac{d y}{d x}=\frac{y^{2} \cot x}{1-y \log \sin x} \end{aligned}$
Hence, it is proved.

Differentiation exercise 10.6 question 6

Answer: $\frac{dy}{dx}=2$
Hint: The value of y is given as infinite series. If a term is deleted from an infinite series, it remains the same in this case.
Given: $y=(\tan x)^{(\tan x)^{(\tan x) \cdot \cdots}}$
Solution:
Here it is given that,
$y=(\tan x)^{(\tan x)^{(\tan x) \cdot \cdots}}$
This can be written as:
$y=(\tan x)^{y}$
Taking log on both sides, we get:
$\log y=\log (\tan x)^{y}$
$\therefore \log y=y\log (\tan x)$ …(1)
Differentiating (1) w.r.t x,
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=y \frac{d}{d x}(\log \tan x)+\log \tan x \cdot \frac{d}{d x}(y) \\ &\frac{1}{y} \frac{d y}{d x}=y \frac{1}{\tan x}\left(\sec ^{2} x\right)+\log \tan x \frac{d y}{d x} \\ &\frac{d y}{d x}\left(\frac{1}{y}-\log \tan x\right)=y \frac{\sec ^{2} x}{\tan x} \\ &\frac{d y}{d x}\left(\frac{1-y \log \tan x}{y}\right)=y \frac{\sec ^{2} x}{\tan x} \\ &\therefore \frac{d y}{d x}=\frac{y^{2}}{1-y \log \tan x} \cdot \frac{\sec ^{2} x}{\tan x} \end{aligned}$
$\text { If } \mathrm{x}=\frac{\pi}{4}, \text { becomes: }$
$\begin{aligned} \left(\frac{d y}{d x}\right)_{x=\left(\frac{\pi}{4}\right)} &=\frac{y^{2}}{1-y \log \tan \left(\frac{\pi}{4}\right)} \cdot \frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{\tan \left(\frac{\pi}{4}\right)} \\ &=\frac{y^{2}(\sqrt{2})^{2}}{1-y \log 1} \\ &=\frac{2 y^{2}}{1-y(0)} \end{aligned}$
$\begin{aligned} \text { Since, }(y) \frac{\pi}{4} &=\tan \left(\frac{\pi}{4}\right)^{\tan \left(\frac{\pi}{4}\right)^{\cdots-\infty}} \\ &=(1)^{\infty} \\ &=1 \end{aligned}$
$\therefore y=1$
$\frac{dy}{dx}=\frac{2\left ( 1 \right )^{2}}{1-0}$
$\frac{dy}{dx}=2$
Hence, it is proved.

Differentiation exercise 10.6 question 8

Answer: $\frac{d y}{d x}=\frac{-y^{2} \tan x}{1-y \log \cos x}$
Hint: The value of y is given as infinite series. If a term is deleted from an infinite series, it remains the same in this case.
Given: $y=\left ( \cos x\right )^{\left ( \cos x \right )^{...\infty }}$
Solution:
Here it is given that,
$y=\left ( \cos x\right )^{\left ( \cos x \right )^{...\infty }}$
This can be written as:
$y=\left ( \cos x\right )^{y}$
Taking log on both sides, we get:
$\begin{aligned} &\log y=\log (\cos x)^{y} \\ &\therefore \log y=y \log (\cos x) \end{aligned}$ …(1)
Differentiating (1) w.r.t x,
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=y \frac{d}{d x}(\log \cos x)+\log \cos x \cdot \frac{d}{d x}(y) \\ &\frac{1}{y} \frac{d y}{d x}=y \frac{1}{\cos x}(-\sin x)+\log \cos x \frac{d y}{d x} \\ &\frac{d y}{d x}\left(\frac{1}{y}-\log \cos x\right)=-y \tan x \\ &\frac{d y}{d x}\left(\frac{1-y \log \cos x}{y}\right)=-y \tan x \\ &\therefore \frac{d y}{d x}=\frac{-y^{2} \tan x}{1-y \log \cos x} \end{aligned}$

Most students mistake solving every single sum from the chapter even if they know the concepts. This is not the suggested method because instead of learning new concepts, they get stuck on the same sort of problem. Instead, students should practice solving until they understand the concept. Once that is done, they can refer to RD Sharma Class 12 Solutions Chapter 10 Ex 10.6 to verify similar problems. This method saves a lot of time and effort and helps them cover a significant portion of the syllabus.

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