Q.If $y=\tan^{-1}\dfrac{x}{1+\sqrt{1-x^2}}+\sin\!\left(2\tan^{-1}\sqrt{\dfrac{1-x}{1+x}}\right)$, then $\dfrac{dy}{dx}=$ (A) $\dfrac{x}{\sqrt{1-x^2}}$ (B) $\dfrac{1-2x}{\sqrt{1-x^2}}$ (C) $\dfrac{1-2x}{2\sqrt{1-x^2}}$ (D) $\dfrac{1-2x^2}{\sqrt{1-x^2}}$

Inverse trigonometric functions ($\sin^{-1}x,\cos^{-1}x,\tan^{-1}x,\cot^{-1}x,\sec^{-1}x,\cosec^{-1}x$) are multi-valued in general, so a principal branch (a restricted domain and range) is fixed for each before differentiating. Each derivative is found by treating $y=$ (inverse trig function of $x$) as $x=$ (trig function of $y$), differentiating implicitly with respect to $y$, and using a Pythagorean identity to express the result back in terms of $x$. The six standard results are: $\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}$, $\dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^2}}$, $\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$, $\dfrac{d}{dx}\cot^{-1}x=-\dfrac{1}{1+x^2}$, $\dfrac{d}{dx}\sec^{-1}x=\dfrac{1}{x\sqrt{x^2-1}}$ (for $x>1$; sign flips for $x<-1$), and $\dfrac{d}{dx}\cosec^{-1}x=-\dfrac{1}{x\sqrt{x^2-1}}$ (for $x>1$; sign flips for $x<-1$). When the argument is a function $f(x)$ rather than pla …