# How do you implicitly differentiate 1=e^(xy)/tany ?

May 5, 2016

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

#### Explanation:

$1 = {e}^{x y} / \tan x$ means $\tan x = {e}^{x y}$. Now differentiating it

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

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

or $x \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{- x y} {\sec}^{2} x - y$

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