Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain Solution R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3 Question 30 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 10 Differentiation  Exercise 10.3 Question 30 Maths Textbook Solution.

Answers (1)

Answer: \frac{dy}{dx}=\frac{3}{1+9x^{2}}-\frac{2}{1+4x^{2}}

Hint:

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constan } \mathrm{t})=0 ; \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}

Given:

\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)

Solution:

Let, y=\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)

y=\tan ^{-1}\left(\frac{3 x-2 x}{1+(3 x)(2 x)}\right)

Since,3x-2x=x

\begin{aligned} &y=\tan ^{-1} 3 x-\tan ^{-1} 2 x \\ &{\left[\text { Since }+\tan x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]} \end{aligned}

Differentiating it with respect tox using chain rule,

\begin{aligned} &\frac{d y}{d x}=\frac{1}{1+\left(3 x^{2}\right)} \frac{d}{d x}(3 x)-\frac{1}{1+(2 x)^{2}} \frac{d}{d x}(2 x) \\ &\left\{\text { Since } \Rightarrow \frac{\mathrm{d} }{\mathrm{d} x}\tan ^{-1}(x)=\frac{1}{1+x^{2}}\right\} \\ &\frac{d y}{d x}=\frac{1}{1+9 x^{2}}(3)-\frac{1}{1+4 x^{2}}(2) \\ &\frac{d y}{d x}=\frac{3}{1+9 x^{2}}-\frac{2}{1+4 x^{2}} \end{aligned}

 

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE launches entrance exam 2025, career guide for parents to plan for students' future
anam-khan

CBSE Class 12 Board Economics Paper Analysis: 'Student-friendly', well balanced with moderate difficulty
anam-khan

CBSE Class 12 Hindi exam 2025 not postponed; special exam for those who miss

Latest Question