# What is the derivative of tan^(-1)(x^2 y^5)?

Apr 17, 2015

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

$\implies \tan u = {x}^{2} {y}^{5}$

By differentiating implicitly with respect to $x$ we have,

$\implies \frac{\mathrm{du}}{\mathrm{dx}} {\sec}^{2} u = \left(2 x\right) {y}^{2} + {x}^{2} \left(5 {y}^{4}\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

$\implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{2 x {y}^{2} + 5 {y}^{4} {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}}{{\sec}^{2} u} = \frac{2 x {y}^{2} + 5 {y}^{4} {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}}{{\tan}^{2} u + 1}$

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

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{2 x {y}^{2} + 5 {y}^{4} {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}}{{\left[\tan \left({\tan}^{-} 1 \left({x}^{2} {y}^{5}\right)\right)\right]}^{2} + 1} = \frac{2 x {y}^{2} + 5 {y}^{4} {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}}{{\left({x}^{2} {y}^{5}\right)}^{2} + 1} = \frac{2 x {y}^{2} + 5 {y}^{4} {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}}{{x}^{4} {y}^{10} + 1}$