# How do you find the derivative of (2t)^5?

Mar 3, 2018

$\frac{d}{\mathrm{dt}} \left[{\left(2 t\right)}^{5}\right] = 160 {t}^{4}$

#### Explanation:

Take out the constant and apply the power rule:

The power rule states: $\frac{d}{\mathrm{dt}} \left[{t}^{n}\right] = n {t}^{n - 1}$

So:

$\frac{d}{\mathrm{dt}} \left[{\left(2 t\right)}^{5}\right] \iff {2}^{5} \cdot \frac{d}{\mathrm{dt}} \left[{t}^{5}\right]$

$= {2}^{5} \cdot \textcolor{red}{5} {t}^{\textcolor{red}{5 - 1}}$

$= {2}^{5} \cdot 5 {t}^{4}$

$= 160 {t}^{4}$