How do you differentiate # f(t)=sin^2(e^(sin^2t)# using the chain rule.?

1 Answer
Dec 20, 2015

Answer:

#f'(t)=2sin(e^(sin^2t))*cos(e^(sin^2t))*e^(sin^2t)*2sintcost#

Explanation:

We will need to use all of the following rules of differentiation :

#d/dx sin(u(x))=cosu*(du)/(dx)#

#d/dxe^(u(x))=e^u*(du)/(dx)#

#d/dx(u(x))^n=n(u(x))^(n-1)*(du)/(dx)#

Applying all these rules in conjunction yields \

#f'(t)=2sin(e^(sin^2t))*cos(e^(sin^2t))*e^(sin^2t)*2sintcost#