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

Apr 23, 2015

Say that $y = \arctan \left({x}^{2} {y}^{5}\right)$ and let's find $\frac{\mathrm{dy}}{\mathrm{dx}}$. We're going to use the implicit differentiation formula below to solve this problem...

$f \left(x\right) \cdot g ' \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + f ' \left(x\right) \cdot g \left(y\right)$

So, let's begin solving the problem...

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

${\sec}^{2} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \cdot 5 {y}^{4} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x \cdot {y}^{5}$

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

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

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \left({x}^{2} {y}^{4} \left({x}^{2} {y}^{6} - 5\right) + 1\right) = 2 x \cdot {y}^{5}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x \cdot {y}^{5}}{{x}^{2} {y}^{4} \left({x}^{2} {y}^{6} - 5\right) + 1}$

To spice things up a little we can find the derivative of this function using partial derivatives...

$z = \arctan \left({x}^{2} {y}^{5}\right)$

$\tan z = {x}^{2} {y}^{5}$ (we hold the function of y as a constant from here)

${\sec}^{2} z \cdot \frac{\partial z}{\partial x} = 2 x {y}^{5}$

$\left({\tan}^{2} z + 1\right) \cdot \frac{\partial z}{\partial x} = 2 x {y}^{5}$

$\left({x}^{4} {y}^{10} + 1\right) \cdot \frac{\partial z}{\partial x} = 2 x {y}^{5}$

$\frac{\partial z}{\partial x} = \frac{2 x {y}^{5}}{{x}^{4} {y}^{10} + 1}$

Alternatively...

$z = \arctan \left({x}^{2} {y}^{5}\right)$

$\tan z = {x}^{2} {y}^{5}$

${\sec}^{2} z \cdot \frac{\partial z}{\partial y} = 5 {x}^{2} {y}^{4}$ (we hold the function of x as a constant from here)

$\left({\tan}^{2} z + 1\right) \cdot \frac{\partial z}{\partial y} = 5 {x}^{2} {y}^{4}$

$\left({x}^{4} {y}^{10} + 1\right) \cdot \frac{\partial z}{\partial y} = 5 {x}^{2} {y}^{4}$

$\frac{\partial z}{\partial y} = \frac{5 {x}^{2} {y}^{4}}{{x}^{4} {y}^{10} + 1}$