# How do you differentiate g(t)=(2/t+t^5)(t^3+1) using the product rule?

Jan 21, 2017

$g ' \left(t\right) = \left(- 2 {t}^{-} 2 + 5 {t}^{4}\right) \left({t}^{3} + 1\right) + \left(2 {t}^{-} 1 + {t}^{5}\right) \cdot 3 {t}^{2}$

#### Explanation:

The product rule says if:
$f \left(x\right) = u \left(x\right) \cdot v \left(x\right)$
then
$f ' \left(x\right) = u ' \left(x\right) \cdot v \left(x\right) + u \left(x\right) \cdot v ' \left(x\right)$

In your equation $g \left(t\right)$
Let
$u \left(t\right) = 2 {t}^{-} 1 + {t}^{5}$ then $u ' \left(t\right) = - 2 {t}^{-} 2 + 5 {t}^{4}$
and
$v \left(t\right) = {t}^{3} + 1$ then $v ' \left(t\right) = 3 {t}^{2}$

So
$g ' \left(t\right) = u ' \left(x\right) \cdot v \left(x\right) + u \left(x\right) \cdot v ' \left(x\right)$

$g ' \left(t\right) = \left(- 2 {t}^{-} 2 + 5 {t}^{4}\right) \left({t}^{3} + 1\right) + \left(2 {t}^{-} 1 + {t}^{5}\right) \cdot 3 {t}^{2}$