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

1 Answer
Oct 30, 2015

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)