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

1 Answer
Sasha P.
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)#

Related questions
See all questions in Chain Rule
Impact of this question
8444 views around the world
You can reuse this answer
Creative Commons License