Q.If $y=\sec(\tan^{-1}x)$ then $\dfrac{dy}{dx}$ at $x=1$ is equal to: (A) $\dfrac12$ (B) $1$ (C) $\dfrac{1}{\sqrt2}$ (D) $\sqrt2$

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 plain $x$, the chain rule multiplies each result by $f'(x)$. Many composite inverse-trig expressions simplify dramatically using a trigonometric substitution ($x=\sin\theta$, $x=\tan\theta$, $x=\sec\theta$, or $x=a\cos2\theta$, guided by the shape of the expression) followed by a double- or triple-angle identity, collapsing the expression to a simple multiple of $\theta$ before differentiating.