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

Question

Christopher Allers · Accepted Answer

#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.

Emilia Aguilar · Answer

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))#

Ezekiel Alexander · Answer

undefined To differentiate $ f(t) = \sin^2(e^{\sin^2t}) $ using the chain rule:

1. Find the derivative of the outer function with respect to its inner function.
$$ \frac{d}{dt}[\sin^2(u)] = 2\sin(u)\cos(u) $$

2. Find the derivative of the inner function with respect to the variable $ t $.
$$ \frac{d}{dt}[e^{\sin^2t}] = e^{\sin^2t} \cdot 2\sin(t)\cos(t) $$

3. Multiply the results from steps 1 and 2.
$$ 2\sin(e^{\sin^2t})\cos(e^{\sin^2t}) \cdot e^{\sin^2t} \cdot 2\sin(t)\cos(t) $$

4. Simplify the expression if necessary.