Question

How do you differentiate `f(x)= ln(tanx-x^2)`?

Answer 1

`(sec^2x-2x)/(tanx-x^2)`

Explanation:

Set `y=lnu, u=tanx-x^2`

`dy/(du)=1/u, (du)/(dx)=sec^2x-2x`

`(dy)/(dx)=(dy)/(du)*(du)/(dx)`

`=1/(tanx-x^2)*sec^2x-2x`

`=(sec^2x-2x)/(tanx-x^2)`