# How do you implicitly differentiate -1=sin(x+y) ?

Nov 24, 2016

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

#### Explanation:

We apply the identity that if $\sin x = y$, then $\arcsin y = x$.

$\arcsin \left(- 1\right) = x + y$

The value of $\arcsin \left(- 1\right)$, or 270˚, is just a constant, so the derivative will be $0$.

$\frac{d}{\mathrm{dx}} \left(\arcsin \left(- 1\right)\right) = \frac{d}{\mathrm{dx}} \left(x + y\right)$

$0 = 1 + \frac{\mathrm{dy}}{\mathrm{dx}}$

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

Hopefully this helps!