Question

Differentiate w.r.t 'x' ,y=${\tan }^{-1}\left(\frac{5-x}{6x^2-5x-3}\right)$

Answer 1

$lett=\frac{5-x}{6x^2-5x-3}$

differentiating both wrt x

$\frac{dt}{dx}=\frac{-(6x^2-5x-3)-(5-x)(12x-5)}{(6x^2-5x-3{)}^2}$

$\frac{dt}{dx}=\frac{6x^2-60x+28}{(6x^2-5x-3{)}^2}$

$\frac{dy}{dx}=\frac{1}{1+\frac{(5-x)}{(6x^2-5x-3}{)}^2}$ . $\frac{d}{dx}\frac{5-x}{6x^2-5x-3}$

$\frac{dy}{dx}=\frac{6x^2-60x+28}{(6x^2-5x-3{)}^2-(x-5{)}^2}$