# What is the derivative of g(t)=(pi)(cos t - 1/t^2)?

Aug 31, 2017

$= \pi \left(\frac{2}{{t}^{3}} - \sin \left(t\right)\right)$

#### Explanation:

remember that $\pi$ is a constant, so you can just mutiply through:

$g \left(t\right) = \pi \cdot \cos \left(t\right) - \frac{\pi}{{t}^{2}} = \pi \cdot \cos \left(t\right) - \pi \cdot {t}^{-} 2$

$\frac{\mathrm{dg}}{\mathrm{dt}} = - \pi \sin \left(t\right) + 2 \pi {t}^{-} 3$

$= \pi \left(\frac{2}{{t}^{3}} - \sin \left(t\right)\right)$