# How do you use implicit differentiation to find dy/dx given sinx+cosy=0?

Dec 21, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

#### Explanation:

First you differentiate the expression, considering that based on the chain rule:

$\frac{\mathrm{dc} o s y}{\mathrm{dx}} = \frac{\mathrm{dc} o s y}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

so that you get:

$\frac{d}{\mathrm{dx}} \left(\sin x + \cos y\right) = \cos x - \sin y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

and resolving for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \frac{x}{\sin} y$

or

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \frac{x}{\sqrt{1 - {\cos}^{2} y}}$

Now we can obtain $\cos y$ from the original equation:

$\cos y = - \sin x$

and then:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \frac{x}{\sqrt{1 - {\sin}^{2} x}} = \cos \frac{x}{\cos} x = 1$

In fact, we can have:

$\cos y = - \sin x$

for any value of $x$ only if $y = x + \frac{\pi}{2} + 2 k \pi$