# How do you differentiate tanx+tany=1?

Feb 20, 2017

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

#### Explanation:

When we differentiate $y$ wrt $x$ we get $\frac{\mathrm{dy}}{\mathrm{dx}}$.

However, we cannot differentiate a non implicit function of $y$ wrt $x$. But if we apply the chain rule we can differentiate a function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

When this is done in situ it is known as implicit differentiation.

We have:

$\tan x + \tan y = 1$

Differentiate wrt $x$:

${\sec}^{2} x + {\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore {\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = - {\sec}^{2} x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - {\sec}^{2} \frac{x}{\sec} ^ 2 y$