Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 chapter Differentiation exercise 10.8 question 20 maths

Explain solution RD Sharma class 12 chapter Differentiation exercise 10.8 question 20 maths

Answers (2)

Answer: \frac{\sqrt{1-x^{2}}}{x\left(1+x^{2}\right)}

Hint:   \tan ^{-1}\left(\frac{a-b}{a+b}\right)=\tan ^{-1} a-\tan ^{-1} b
Given:  \tan ^{-1}\left(\frac{1-x}{1+x}\right) \text { w.r.t } \sqrt{1-x^{2}}

            -1<x<1

Explanation:

\tan ^{-1}\left(\frac{1-x}{1+x}\right)=u

\begin{aligned} &u=\tan ^{-1}(1)-\tan ^{-1} x \\\\ &u=\frac{\pi}{4}-\tan ^{-1} x \\\\ &\frac{d u}{d x}=0-\frac{1}{1+x^{2}}=-\frac{1}{1+x^{2}} \end{aligned}

\begin{aligned} &v=\sqrt{1-x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-x^{2}}}(-2 x) \\\\ &\frac{d v}{d x}=\frac{-x}{\sqrt{1-x^{2}}} \end{aligned}

\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{-\frac{1}{1+x^{2}}}{\frac{-x}{\sqrt{1-x^{2}}}}=\frac{\sqrt{1-x^{2}}}{x\left(1+x^{2}\right)}

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

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

Answer: \frac{\sqrt{1-x^{2}}}{x\left(1+x^{2}\right)}

Hint:   \tan ^{-1}\left(\frac{a-b}{a+b}\right)=\tan ^{-1} a-\tan ^{-1} b
Given:  \tan ^{-1}\left(\frac{1-x}{1+x}\right) \text { w.r.t } \sqrt{1-x^{2}}

            -1<x<1

Explanation:

\tan ^{-1}\left(\frac{1-x}{1+x}\right)=u

\begin{aligned} &u=\tan ^{-1}(1)-\tan ^{-1} x \\\\ &u=\frac{\pi}{4}-\tan ^{-1} x \\\\ &\frac{d u}{d x}=0-\frac{1}{1+x^{2}}=-\frac{1}{1+x^{2}} \end{aligned}

\begin{aligned} &v=\sqrt{1-x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-x^{2}}}(-2 x) \\\\ &\frac{d v}{d x}=\frac{-x}{\sqrt{1-x^{2}}} \end{aligned}

\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{-\frac{1}{1+x^{2}}}{\frac{-x}{\sqrt{1-x^{2}}}}=\frac{\sqrt{1-x^{2}}}{x\left(1+x^{2}\right)}

Posted by

infoexpert26

View full answer

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question