Question

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

Answer 1

`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?