Question

Differentiate w.r.t. $x$ in ${\tan }^{-1}\left(\frac{5x}{1-6x^2}\right)$

Answer 1

Let,

$y={\tan }^{-1}\left(\frac{5x}{1-6x^2}\right)$

$={\tan }^{-1}\left(\frac{3x+2x}{1-3x\times 2x}\right)$

$\Rightarrow y={\tan }^{-1}3x+{\tan }^{-1}2x$

$\frac{dy}{dx}=\frac{1}{1+(3x{)}^2}\frac{d}{dx}\left(3x\right)+\frac{1}{1+(2x{)}^2}\frac{d}{dx}\left(2x\right)$

$=\frac{3}{1+9x^2}+\frac{2}{1+4x^2}$