# How do you determine dy/dx if y=x+tan(xy)?

May 23, 2018

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

#### Explanation:

we get by the chain rule:
$y ' = 1 + \frac{1}{\cos} ^ 2 \left(x y\right) \left(y + x y '\right)$

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

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

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

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

Multiplying by ${\cos}^{2} \left(x y\right)$ over ${\cos}^{2} \left(x y\right)$ we get

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