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

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Answer 1 · 2016-03-16T16:29:51

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

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 · 2016-03-16T19:04:26

Please see the explanation below.

#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))#