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

2 Answers
Mar 16, 2016

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.

Mar 16, 2016

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))