Question

What is the derivative of `y = ln(sin^2(theta))`?

Answer 1

`Cot^2theta`

Explanation:

Using the chain rule we have:

`F(x)'= f'(g(x))*g'(x)`

`y'=ln'(sin^2(theta))*(cos^2(theta))`
`1/sin^2theta *Cos^2theta`

`Cot^2theta`