Find $\frac{dy}{dx}$ where $y=\sin^{-1}(\frac{1-x^2}{1+x^2})$
Answer & explanation
Correct answer: option 3
$y=\sin^{-1}(\frac{1-x^2}{1+x^2})=\frac{π}{2}-\cos^{-1}(\frac{1-x^2}{1+x^2})$
$y=\frac{π}{2}-2\tan^{-1}x$
so $\frac{dy}{dx}=\frac{-2}{1+x^2}$
Find $\frac{dy}{dx}$ where $y=\sin^{-1}(\frac{1-x^2}{1+x^2})$
Correct answer: option 3
$y=\sin^{-1}(\frac{1-x^2}{1+x^2})=\frac{π}{2}-\cos^{-1}(\frac{1-x^2}{1+x^2})$
$y=\frac{π}{2}-2\tan^{-1}x$
so $\frac{dy}{dx}=\frac{-2}{1+x^2}$