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

May 26, 2016

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

#### Explanation:

We have to find derivative of ${\tan}^{- 1} \left({x}^{2} {y}^{5}\right)$. For this we use the concept of chain rule for implicit differentiation as well as product rule..

Hence $\frac{d}{\mathrm{dx}} \left({\tan}^{- 1} \left({x}^{2} {y}^{5}\right)\right)$

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

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