# How do you differentiate g(w) = (w^n - 5^n) /n ?

Oct 11, 2015

Assuming that $n$ is a constant, $g ' \left(w\right) = {w}^{n - 1}$

#### Explanation:

Assuming that $n$ is a constant,

$g \left(w\right) = \frac{{w}^{n} - {5}^{n}}{n} = \frac{1}{n} \left({w}^{n} - {5}^{n}\right)$

and ${5}^{n}$ is constant, so its derivative is $0$.

$g ' \left(w\right) = \frac{1}{n} \frac{d}{\mathrm{dw}} \left({w}^{n} - {5}^{n}\right)$

$= \frac{1}{n} \left(n {w}^{n - 1} - 0\right) = {w}^{n - 1}$