# How to find the derivative of the following function: f(t)= 1/ 2t^3? Thank you

Feb 11, 2017

$f ' \left(t\right) = - \frac{3}{2 {t}^{4}}$

#### Explanation:

$f \left(t\right) = \frac{1}{2 {t}^{3}}$.

Since ${a}^{-} b = \frac{1}{a} ^ b$, we can rewrite $\frac{1}{t} ^ 3 = {t}^{-} 3$

$f \left(t\right) = \frac{1}{2} {t}^{-} 3$.

Using the power rule, which states that $\left({x}^{n}\right) ' = n {x}^{n - 1}$, we get:

$f ' \left(t\right) = \frac{1}{2} \cdot \left({t}^{-} 3\right) ' = \frac{1}{2} \cdot \left(- 3 {t}^{-} 4\right) = - \frac{3}{2} {t}^{-} 4 = - \frac{3}{2 {t}^{4}}$