How do you find the derivative of #W(u)=e^(u^2)*tan(3sqrtu)#?

1 Answer
Dec 10, 2017

Answer:

#W'(u) = (3e^(u^2))/(2root2u)sec^2(3root2u) + 2ue^(u^2)tan(3e^(1/2))#

Explanation:

The answer is, it is very annoying.

The formula for product rule is
#f(x) = uv#
#f'(x) = udv + vdu#
which seems simple enough.

Since there are clearly two terms in your function, lets set one equal to #u# and on equal to #v#.

#W(u) = e^(u^2) * tan(3u^(1/2))#

Our first dilema is that #u# is already used in the function, so instead I'll use the letter #z#. Then, I'll have to change the product rule formula.

#W(u) = zv#
#W'(u) = zdv + vdz#

So now,

#z = e^(u^2)#
and
#v = tan(3u^(1/2))#

Looking at the product rule, we need four things:
#z#, #v#, #dz#, and #dv#.
We already have #z# and #v# so lets find #dz# and #dv#.

This is where it gets annoying.

#z = e^(u^2)#
so
#dz = 2ue^(u^2)#

#v = tan(3u^(1/2))#
so
#dv = 3/2u^(-1/2)sec^2(3u^(1/2))#

If you need further explanation of how I got these, let me know.

Now we just plug these variables into the formula.

#W'(u) = zdv + vdz#
#W'(u) = e^(u^2) * 3/2u^(-1/2)sec^2(3u^(1/2)) + tan(3u^(1/2)) * 2ue^(u^2)#

Simplify it to get

#W'(u) = (3e^(u^2))/(2root2u)sec^2(3root2u) + 2ue^(u^2)tan(3e^(1/2))#

See, wasn't that annoying?