Question

Differentiate ${\tan }^{-1}\frac{x}{\sqrt{a^2-x^2}}$, with respect to $x$.

Answer 1

Let $y={\tan }^{-1}\frac{x}{\sqrt{a^2-x^2}}$
$x=a\sin \theta$ (Let)
$\therefore \theta ={\sin }^{-1}\frac{x}{a}....\left(1\right)$
$y={\tan }^{-1}\frac{a\sin \theta }{\sqrt{a^2-a^2{\sin }^2\theta }}$
$={\tan }^{-1}\frac{a\sin \theta }{\sqrt{a^2{\cos }^2\theta }}={\tan }^{-1}\frac{a\sin \theta }{a\cos \theta }$
$={\tan }^{-1}\tan \theta$
$y=\theta$
$y=\theta ={\sin }^{-1}\frac{x}{a}\left(by\right(1\left)\right)$
differentiate with respect to $x$
$\frac{d}{dx}y=\frac{d}{dx}{\sin }^{-1}\frac{x}{a}$
$=\frac{1}{a}\times \frac{1}{\sqrt{1-\frac{x^2}{a^2}}}$
$=\frac{1}{a}\times \frac{a}{\sqrt{a^2-x^2}}=\frac{1}{\sqrt{a^2-x^2}}$
$\frac{d}{dx}{\tan }^{-1}\frac{x}{\sqrt{a^2-x^2}}=\frac{1}{\sqrt{a^2-x^2}}$