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

##### 1 Answer

#### 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

So, where

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

For the leftover derivative, the overriding function is the sine function. Where

Where

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

Now the overriding function is the power of

Here,

#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

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