# How do you differentiate s=(t^-1+t^-2)/t^-3?

Nov 21, 2016

$s ' \left(t\right) = 2 t + 1$

#### Explanation:

First rewrite the function:

$s \left(t\right) = \frac{\frac{1}{t} + \frac{1}{t} ^ 2}{\frac{1}{t} ^ 3} = {t}^{3} \left(\frac{1}{t} + \frac{1}{t} ^ 2\right) = {t}^{2} + t$

Through the product rule, which states that $\frac{d}{\mathrm{dt}} \left({t}^{n}\right) = n {t}^{n - 1}$, we see that:

$s ' \left(t\right) = 2 {t}^{2 - 1} + 1 {t}^{1 - 1} = 2 {t}^{1} + 1 {t}^{0} = 2 t + 1$