Question

How do you find the derivative of `f(t)=sin^2[e^(sin^2)t]` using the chain rule?

Answer 1

`f'(t)=e^(sin^2t) * sin2t * sin2e^(sin^2t)`

Explanation:

`f'(t)=(sin^2e^(sin^2t))' = 2sine^(sin^2t) * (sine^(sin^2t))'`

`f'(t)=2sine^(sin^2t) * cose^(sin^2t) * (e^(sin^2t))'`

`f'(t)=2sine^(sin^2t) * cose^(sin^2t) * e^(sin^2t) *(sin^2t)'`

`f'(t)=2sine^(sin^2t) * cose^(sin^2t) * e^(sin^2t) * 2sint * (sint)'`

`f'(t)=2sine^(sin^2t) * cose^(sin^2t) * e^(sin^2t) * 2sint * cost`

`f'(t)=e^(sin^2t) * sin2t * sin2e^(sin^2t)`