If $y = an^{-1}\left(\dfrac{2x}{1-x^2} ight)$, then $\dfrac{dy}{dx}…

Question

If $y = 	an^{-1}\left(\dfrac{2x}{1-x^2}ight)$, then $\dfrac{dy}{dx}$ equals:

Accepted Answer

A. $\dfrac{2}{1+x^2}$

Recognise that $	an^{-1}(2x/(1-x^2)) = 2	an^{-1}x$ (double-angle identity, $|x| < 1$).
So $dy/dx = 2/(1+x^2)$.
Direct quotient-rule differentiation gives the same answer.

Answer

$\dfrac{-2}{1+x^2}$

Answer

$\dfrac{1}{\sqrt{1-x^2}}$

Answer

$\dfrac{1}{1+x^2}$

If $y = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$, then $\dfrac{dy}{dx}$ equals:

Keep practising Unit 2: Differentiation and Fundamental Derivative Rules