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

Nov 10, 2016

Start by expanding $\sin \left(x + y\right)$ using the sum and difference identity.

$\sin \left(x + y\right) = x y$

$\sin x \cos y + \cos x \sin y = x y$

By the product rule:

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

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

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

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

By the formula $\cos A \cos B - \sin A \sin B = \cos \left(A + B\right)$, we have:

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

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