# How do you find the derivative of y = sin(x+y)?

Jul 28, 2016

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

#### Explanation:

You simply differentiate both sides with respect to $x$. The left side would simply give you $\frac{\mathrm{dy}}{\mathrm{dx}}$. For the right side, however, you must make use of the chain rule for derivatives of composite functions (functions of functions). Thus

$\frac{d}{\mathrm{dx}} \left(\sin \left(x + y\right)\right) = \cos \left(x + y\right) \times \frac{d}{\mathrm{dx}} \left(x + y\right) = \cos \left(x + y\right) \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Thus, we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(x + y\right) \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

We can easily solve this for the quantity $\frac{\mathrm{dy}}{\mathrm{dx}}$:

(1-(cos(x+y)) dy/dx = cos(x+y) implies dy/dx= cos(x+y)/{1-cos(x+y)}