Question

Differentiate: $y=\frac{1}{ab}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)$ with respect to $x$.

Answer 1

Let ,$y=\frac{1}{ab}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)$

Differentiate with respect to x

$\frac{dy}{dy}=\frac{1}{ab}\frac{d}{dx}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)$

$=\frac{1}{ab}\times \frac{1}{1+{\left(\frac{b}{a}\tan x\right)}^2}\frac{d}{dx}\left(\frac{b}{a}\tan x\right)$

$=\frac{1}{ab}\times \frac{1}{1+{\left(\frac{b}{a}\tan x\right)}^2}\frac{b}{a}\frac{d}{dx}\left(\tan x\right)$

$=\frac{1}{ab}\times \frac{a^2}{a^2+{\left(b\tan x\right)}^2}\frac{b}{a}{\sec }^{2}x$

$\frac{dy}{dx}=\frac{1}{a^2+b^2{\tan }^{2}x}{\sec }^{2}x$

$\frac{dy}{dx}=\frac{1}{a^2+b^2{\tan }^{2}x}{\sec }^{2}x$

$\frac{dy}{dx}=\frac{{\sec }^{2}x}{a^2+b^2{\tan }^2\theta }$