What is the derivative of #f(t) = sin^2 (e^(sin^2t))#?

1 Answer
Jan 24, 2017

#f'(t)=sin(2e^(sin^2t))e^(sin^2t)(sin(2t))#

Explanation:

We will peel off functions one by one using the chain rule. The overriding function is the squared function.

The chain rule states that the derivative of #g(t)^2=2g(t)^1*g'(t)#.

So, where #g(t)=sin(e^(sin^2t))#:

#f'(t)=2sin(e^(sin^2t))*d/dtsin(e^(sin^2t))#

For the leftover derivative, the overriding function is the sine function. Where #sin(h(t))#, the derivative is #cos(h(t))*h'(t)#.

Where #h(t)=e^(sin^2t)#, this becomes:

#f'(t)=2sin(e^(sin^2t))*cos(e^(sin^2t))*d/dte^(sin^2t)#

Now the overriding function is the power of #e#. For a function #e^(j(t))#, the derivative through the chain rule is #e^(j(t))*j'(t)#.

Here, #j(t)=sin^2t#, so:

#f'(t)=2sin(e^(sin^2t))cos(e^(sin^2t))*e^(sin^2t)*d/dtsin^2t#

We again have a squared function. Following the method before:

#f'(t)=2sin(e^(sin^2t))cos(e^(sin^2t))e^(sin^2t)*2sint*d/dtsint#

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

Using #2sinthetacostheta=sin2theta# this can be rewritten as:

#f'(t)=sin(2e^(sin^2t))e^(sin^2t)(sin(2t))#