# How do you differentiate sinx+siny=1?

Jun 29, 2018

The answer is $= - \cos \frac{x}{- {\sin}^{2} x + 2 \sin x}$

#### Explanation:

This is an implicit differentiation

$\sin x + \sin y = 1$

$\sin y = 1 - \sin x$

$\cos y = \sqrt{1 - {\left(1 - \sin x\right)}^{2}} = - {\sin}^{2} x + 2 \sin x$

Differentiating wrt $x$

$\cos x + \frac{\mathrm{dy}}{\mathrm{dx}} \cos y = 0$

$\implies$, $\frac{\mathrm{dy}}{\mathrm{dx}} \cos y = - \cos x$

$\implies$, $\frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \frac{x}{\cos} y$

$\implies$, $\frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \frac{x}{- {\sin}^{2} x + 2 \sin x}$