Question

$y=\sin \left(2{\sin }^{-1}x\right),\frac{dy}{dx}=$

• A. $2\sqrt{\left(\frac{1-y^2}{1-x^2}\right)}.$
• B. $2\sqrt{\left(\frac{1+y^2}{1-x^2}\right)}.$
• C. $2\sqrt{\left(\frac{1-y^2}{1+x^2}\right)}.$
• D. $2\sqrt{\left(\frac{1+y^2}{1+x^2}\right)}.$

Answer 1

The correct option is A $2\sqrt{\left(\frac{1-y^2}{1-x^2}\right)}.$

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

$\therefore {\sin }^{-1}y=2{\sin }^{-1}x.$

Differentiate both sides w.r.t. $x$,

$\frac{1}{\sqrt{(1-y^2)}}\frac{dy}{dx}=2.\frac{1}{\sqrt{(1-x^2)}}.$

$\therefore \frac{dy}{dx}=2\sqrt{\left(\frac{1-y^2}{1-x^2}\right)}.$