Question

Differentiate the following functions w.r.t.x
${\tan }^{-1}\left(\frac{\sqrt{1+a^2{x}^2}}{ax}\right)$

Answer 1

Let $y=ta{n}^{-1}\left(\frac{\sqrt{1+a^2{x}^2}}{ax}\right)$

$tany=\frac{\sqrt{1+a^2{x}^2}}{ax}\Rightarrow ta{n}^{2}y=\frac{1+a^2{x}^2}{a^2{x}^2}$

$a^2{x}^{2}ta{n}^{2}y=1+a^2{x}^2$ by squaring both sides

$a^2{x}^2(ta{n}^{2}y-1)=1$

$a^2[x^{2}2tanyse{c}^{2}y\frac{dy}{dx}+2x(ta{n}^{2}y-1\left)\right]=0$

$\Rightarrow 2a^2{x}^{2}tanyse{c}^{2}y.\frac{dy}{dx}=2x(1-ta{n}^{2}y)$

$\Rightarrow a^{2}x\frac{\sqrt{1+a^2{x}^2}}{ax}se{c}^{2}y\frac{dy}{dx}=1-\frac{1+a^2{x}^2}{a^2{x}^2}$

$\Rightarrow a\sqrt{1+a^2{x}^2}(1+ta{n}^{2}y)\frac{dy}{dx}=\frac{a^2{x}^2-1-a^2{x}^2}{a^2{x}^2}$

$\Rightarrow (1+\frac{1+a^2{x}^2}{a^2{x}^2})\frac{dy}{dx}=\frac{-1}{a^2{x}^2\times a\sqrt{1+a^2{x}^2}}$

$\Rightarrow \frac{2a^2{x}^2+1}{a^2{x}^2}\frac{dy}{dx}=\frac{-1}{a^2{x}^2\times a\sqrt{1+a^2{x}^2}}$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{a^2{x}^2\times a\sqrt{1+a^2{x}^2}}$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{a^2{x}^2\times a\sqrt{1+a^2{x}^2}}$