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

Dec 10, 2017

$W ' \left(u\right) = \frac{3 {e}^{{u}^{2}}}{2 \sqrt[2]{u}} {\sec}^{2} \left(3 \sqrt[2]{u}\right) + 2 u {e}^{{u}^{2}} \tan \left(3 {e}^{\frac{1}{2}}\right)$

#### Explanation:

The answer is, it is very annoying.

The formula for product rule is
$f \left(x\right) = u v$
$f ' \left(x\right) = u \mathrm{dv} + v \mathrm{du}$
which seems simple enough.

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

$W \left(u\right) = {e}^{{u}^{2}} \cdot \tan \left(3 {u}^{\frac{1}{2}}\right)$

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.

$W \left(u\right) = z v$
$W ' \left(u\right) = z \mathrm{dv} + v \mathrm{dz}$

So now,

$z = {e}^{{u}^{2}}$
and
$v = \tan \left(3 {u}^{\frac{1}{2}}\right)$

Looking at the product rule, we need four things:
$z$, $v$, $\mathrm{dz}$, and $\mathrm{dv}$.
We already have $z$ and $v$ so lets find $\mathrm{dz}$ and $\mathrm{dv}$.

This is where it gets annoying.

$z = {e}^{{u}^{2}}$
so
$\mathrm{dz} = 2 u {e}^{{u}^{2}}$

$v = \tan \left(3 {u}^{\frac{1}{2}}\right)$
so
$\mathrm{dv} = \frac{3}{2} {u}^{- \frac{1}{2}} {\sec}^{2} \left(3 {u}^{\frac{1}{2}}\right)$

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

Now we just plug these variables into the formula.

$W ' \left(u\right) = z \mathrm{dv} + v \mathrm{dz}$
$W ' \left(u\right) = {e}^{{u}^{2}} \cdot \frac{3}{2} {u}^{- \frac{1}{2}} {\sec}^{2} \left(3 {u}^{\frac{1}{2}}\right) + \tan \left(3 {u}^{\frac{1}{2}}\right) \cdot 2 u {e}^{{u}^{2}}$

Simplify it to get

$W ' \left(u\right) = \frac{3 {e}^{{u}^{2}}}{2 \sqrt[2]{u}} {\sec}^{2} \left(3 \sqrt[2]{u}\right) + 2 u {e}^{{u}^{2}} \tan \left(3 {e}^{\frac{1}{2}}\right)$

See, wasn't that annoying?