Question

What is the derivative of `x sin y + y cos x = 1`?

Answer 1

`y'=(ysinx-siny)/(xcosy+cosx)`

Explanation:

Differentiate the equation with respect to `x`:

`d/(dx) (xsiny+ycosx) = 0`

`xy'cosy+siny + y'cosx -ysinx = 0`

`y'(xcosy+cosx) = ysinx-siny`

`y'=(ysinx-siny)/(xcosy+cosx)`