Search any question & find its solution
Question:
Answered & Verified by Expert
Find $\frac{d y}{d x}$ in the following :
$y=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right),-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$
$y=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right),-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$
Solution:
2510 Upvotes
Verified Answer
$\begin{aligned} y &=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), \text { put } x=\sin \theta \\ y &=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^2 \theta}\right) \\ &=\sin ^{-1}(2 \sin \theta \cos \theta)=\sin ^{-1}(\sin 2 \theta)=2 \theta \\ y &=2 \sin ^{-1} x \therefore \frac{d y}{d x}=\frac{2}{\sqrt{1-x^2}} \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.