# How do you differentiate  x/e^(3x)?

##### 1 Answer
Mar 13, 2016

Apply the quotient rule to get $\frac{1 - 3 x}{{e}^{3 x}}$.

#### Explanation:

The quotient rule says that, for two functions of $x$, call them $u$ and $v$,
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$
In this problem, $u = x$ and $v = {e}^{3 x}$. Step 1 is to take the derivative of $u$ and $v$ to make the calculations easier:
$u = x$
$u ' = 1$

$v = {e}^{3 x}$
$v ' = 3 {e}^{3 x}$
We can now make the substitutions:
$\frac{d}{\mathrm{dx}} \left(\frac{x}{e} ^ \left(3 x\right)\right) = \frac{\left(x\right) ' \left({e}^{3 x}\right) - \left(x\right) \left({e}^{3 x}\right) '}{{\left({e}^{3 x}\right)}^{2}}$
$\frac{d}{\mathrm{dx}} \left(\frac{x}{e} ^ \left(3 x\right)\right) = \frac{{e}^{3 x} - 3 x {e}^{3 x}}{{\left({e}^{3 x}\right)}^{2}}$

And finish off with a little algebra:
$\frac{d}{\mathrm{dx}} \left(\frac{x}{e} ^ \left(3 x\right)\right) = \frac{{e}^{3 x} \left(1 - 3 x\right)}{{\left({e}^{3 x}\right)}^{2}}$
$\frac{d}{\mathrm{dx}} \left(\frac{x}{e} ^ \left(3 x\right)\right) = \frac{1 - 3 x}{{e}^{3 x}}$