Question

How do you differentiate `xcosy+ycos x=1`?

Answer 1

Differentiate `xcosy+ycos x=1`

I assume that you want `(dy)/(dx)`, which is

`(dy)/(dx)=(ysinx-cosy)/(cosx-xsiny)`.

Method:
Since we can't (or don't want to) solve for `y`, leave the function(s) implicit and use implicit differentiation. (An application of the chain rule.)

Given: `xcosy+ycos x=1`
We know: `d/(dx)(xcosy+ycos x)= d/(dx)(1)`
So, using the product rule twice on the left:
`[(dx)/(dx)cosy+xd/(dx)(cosy)]+[(dy)/(dx)cos x+yd/(dx)(cosx)]=0`

Thus, we see that:

`cosy-xsiny (dy)/(dx)+(dy)/(dx)cosx-ysinx=0`

Solve algebraically for `(dy)/(dx)`, to get:

`(dy)/(dx)=(ysinx-cosy)/(cosx-xsiny)`.