Question

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

Answer 1

`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))`