Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Multiple choice question 10

Provide solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Multiple choice question 10

Answers (1)

Answer:

        \frac{-2}{1+x^{2}}

Hint:

        Differentiate y function w.r.t x

Given:

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

Solution:  

Let

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

Differentiate y w.r.t x then

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

               =\frac{1}{\sqrt{1-\left(\frac{1-x^{2}}{1+x^{2}}\right)^{2}}}\left\{\frac{\left(1+x^{2}\right) \frac{d}{d x}\left(1-x^{2}\right)-\left(1-x^{2}\right) \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right\}        \left[\because \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}, \frac{d}{d x}\left(\frac{u}{v}\right)=v \frac{d u}{d x}-u \frac{d v}{d x}\right]

               =\frac{1}{\sqrt{\frac{\left(1+x^{2}\right)^{2}-\left(1-x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}}}}\left[\frac{\left(1+x^{2}\right)\left\{\frac{d}{d x}(1)-\frac{d}{d x}\left(x^{2}\right)\right\}-\left(1-x^{2}\right)\left\{\frac{d}{d x}(1)+\frac{d}{d x}\left(x^{2}\right)\right\}}{\left(1+x^{2}\right)^{2}}\right]                \left[\because \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x)\right]

               =\frac{\sqrt{\left(1+x^{2}\right)^{2}}}{\sqrt{\left(1+x^{4}+2 x^{2}\right)-\left(1+x^{4}-2 x^{2}\right)}}\left[\frac{\left(1+x^{2}\right)(0-2 x)-\left(1-x^{2}\right)(0+2 x)}{\left(1+x^{2}\right)^{2}}\right]

                                                                                                                \left[\frac{d}{d x}\left(x^{n}\right)=n x^{n-1},\left(a^{2}+b^{2}+2 a b\right)=(a+b)^{2} \&(a-b)^{2}=a^{2}+b^{2}-2 a b\right]

               =\frac{\left(1+x^{2}\right)}{\sqrt{1+x^{4}+2 x^{2}-1-x^{4}+2 x^{2}}}\left[\frac{-2 x-2 x^{3}-\left(2 x-2 x^{3}\right)}{\left(1+x^{2}\right)^{2}}\right]

               =\frac{1+x^{2}}{\sqrt{4 x^{2}}}\left[\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{\left(1+x^{2}\right)^{2}}\right]

               \begin{aligned} &\left.=\frac{1+x^{2}}{\sqrt{(2 x)^{2}}\left[\frac{-4 x}{\left(1+x^{2}\right)^{2}}\right.}\right]=\frac{\left(1+x^{2}\right)}{2 x} \times \frac{-4 x}{\left(1+x^{2}\right)^{2}}=\frac{-2}{\left(1+x^{2}\right)} \\\\ &\therefore \frac{d y}{d x}=\frac{-2}{1+x^{2}} \end{aligned}

 

 

Posted by

infoexpert26

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

NSUI stages protest outside CBSE headquarters, demands rollback of OSM
anam-khan

CBSE revaluation payment system faces malicious attack; IIT experts, PSU banks step in
anam-khan

From next year, CBSE Class 12 answer sheets on Digilocker: Education ministry

Latest Question