Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The differential coefficient of  <img alt="\tan ^{-1} \frac{2 x}{1-x^2}

The differential coefficient of  \tan ^{-1} \frac{2 x}{1-x^2} \text { w.r.t. } \sin ^{-1} \frac{2 x}{1+x^2} \text { is }

Option: 1

1


Option: 2

-1


Option: 3

0


Option: 4

None of these


Answers (1)

best_answer

Let y_1=\tan ^{-1} \frac{2 x}{1-x^2} \text { and } y_2=\sin ^{-1} \frac{2 x}{1+x^2}

Putting x = tanθ

\therefore y_1=\tan ^{-1} \tan 2 \theta=2 \theta=2 \tan ^{-1} x \text { and } y_2=\sin ^{-1} \sin 2 \theta=2 \tan ^{-1} x

Again \frac{d y_1}{d x}=\frac{d}{d x}\left[2 \tan ^{-1} x\right]=\frac{2}{1+x^2}...(1)

and \frac{d y_2}{d x}=\frac{d}{d x}\left[2 \tan ^{-1} x\right]=\frac{2}{1+x^2}.....(2)

Hence \frac{d y_1}{d y_2}=1

 

Posted by

Riya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question