Question

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

Answer 1

`dy/dt=cos^2t(e^(sin^2)t)*e^(sin^2)t*cos^2t`

Explanation:

So, we got three functions here:

`sin^2(e^(sin^2)t)`

`e^(sin^2)t`

and

Let `y=sin^2(e^(sin^2)t)`

differentiating w.r.t. `t`

`dy/dt=d/dtsin^2(e^(sin^2)t)`

`dy/dt=cos^2t(e^(sin^2)t)*d/dte^(sin^2)t`

`dy/dt=cos^2t(e^(sin^2)t)*e^(sin^2)t*d/dtsin^2t`

`dy/dt=cos^2t(e^(sin^2)t)*e^(sin^2)t*cos^2t`

This will be the differentiated function.

Answer 2

Please see the explanation below.

Explanation:

`f(t) = sin^2(e^(sin^2t))` has the form

`sin^2(u)` which is also `(sin(u))^2`

So we need the derivative of a square and we'll need the chain rule.

`d/dt((sin(u))^2) = 2(sin(u))*d/dt(sin(u))`

`= 2sin(u)cos(u) d/dt(u)`

In this problem, `u = e^(sin^2t)`

So,
`(du)/dt = e^(sin^2t)*d/dt(sin^2t)`

`= e^(sin^2t)*[2sintcost]`

Combining all of this into one calculation:

`f'(t) = 2sin(e^(sin^2t))cos(e^(sin^2t))e^(sin^2t)*[2sintcost]`

`= 4sintcost * e^(sin^2t)sin(e^(sin^2t))cos(e^(sin^2t))`

Because `2sinthetacostheta = sin(2theta)`, this couls also be written as

`= [2sintcost] * [e^(sin^2t)][2sin(e^(sin^2t))cos(e^(sin^2t))]`

`= sin(2t) e^(sin^2t)sin(2e^(sin^2t))`