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

##### 1 Answer
Apr 17, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x y + 2 {x}^{3} {y}^{3} - y}{x - {x}^{2} - {x}^{4} {y}^{2}}$

#### Explanation:

We will be differentiating implicitly. On the left hand side, we will use the chain rule in regards to the inverse tangent function:

$\frac{d}{\mathrm{dx}} \left(\arctan \left(u\right)\right) = \frac{u '}{1 + {u}^{2}}$

Also, note that the product rule will be used for $\frac{d}{\mathrm{dx}} \left({x}^{2} y\right)$ and, eventually, $\frac{d}{\mathrm{dx}} \left(x y\right)$.

Differentiating gives:

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

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

$\frac{y + x \frac{\mathrm{dy}}{\mathrm{dx}}}{1 + {x}^{2} {y}^{2}} = 2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$

From here, just algebra your way through to an equation solved for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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

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

$x \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{2} + {x}^{4} {y}^{2}\right) = 2 x y + 2 {x}^{3} {y}^{3} - y$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(x - {x}^{2} - {x}^{4} {y}^{2}\right) = 2 x y + 2 {x}^{3} {y}^{3} - y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x y + 2 {x}^{3} {y}^{3} - y}{x - {x}^{2} - {x}^{4} {y}^{2}}$