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

Question

Isaiah Allen · Accepted Answer

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

The response is that it annoys me greatly.

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 the point of annoyance.

#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))#

Tell me if you require any additional details on how I obtained these.

We simply need to enter these variables into the formula now.

#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)#

Make it simpler to understand

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

See, wasn't that bothersome?

Samantha Adkison · Answer

undefined To find the derivative of $ W(u) = e^{u^2} \cdot \tan(3\sqrt{u}) $, we use the product rule and chain rule.

$$ W'(u) = e^{u^2} \cdot \sec^2(3\sqrt{u}) \cdot \frac{d}{du}(u^2) + 2u \cdot e^{u^2} \cdot \tan(3\sqrt{u}) $$

$$ W'(u) = e^{u^2} \cdot \sec^2(3\sqrt{u}) \cdot (2u) + 2u \cdot e^{u^2} \cdot \tan(3\sqrt{u}) $$