# How do you differentiate sinx+cosy=sinxcosy?

Jun 23, 2015

I found:
(dy)/(dx)=(cos(x)(cos(y)-1))/(sin(y)(sin(x)-1)

#### Explanation:

You need to remember the $y$ will be a function of $x$ and differentiate accordingly, so you have:

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

where on the right I used the Product Rule;

now collect $\frac{\mathrm{dy}}{\mathrm{dx}}$:

(dy)/(dx)=(-cos(x)+ cos(x)cos(y))/(-sin(y)+sin(x)sin(y))=(cos(x)(cos(y)-1))/(sin(y)(sin(x)-1)