If $y = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$, then $\dfrac{dy}{dx}$ equals:
A$\dfrac{2}{1+x^2}$
B$\dfrac{-2}{1+x^2}$
C$\dfrac{1}{\sqrt{1-x^2}}$
D$\dfrac{1}{1+x^2}$
Answer & Solution
Correct answer: A. $\dfrac{2}{1+x^2}$
Recognise that $\tan^{-1}(2x/(1-x^2)) = 2\tan^{-1}x$ (double-angle identity, $|x| < 1$). So $dy/dx = 2/(1+x^2)$. Direct quotient-rule differentiation gives the same answer.
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