Question

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

Answer 1

`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`