Question

How do you implicitly differentiate `9=sin^2x-cos^2y`?

Answer 1

`dy/dx=-(sinxcosx)/(sinycosy)`

Explanation:

Recall that when implicitly differentiating, any derivatives of terms with `y` will spit out a `dy/dx` term thanks to the chain rule.

`d/dx[9=sin^2x-cos^2y]`

`0=2sinx*d/dx[sinx]-2cosy*d/dx[cosy]`

`0=2sinxcosx-2cosy(-siny)dy/dx`

`dy/dx=-(sinxcosx)/(sinycosy)`