Question

How do you differentiate `sinx+cosy=sinxcosy`?

Answer 1

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(x)-sin(y)(dy)/(dx)=cos(x)cos(y)-sin(x)sin(y)(dy)/(dx)`

where on the right I used the Product Rule;

now collect `(dy)/(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)`