# How do you find the derivative of (t^3+1)^100 using the chain rule?

Nov 24, 2015

$300 {t}^{2} {\left({t}^{3} + 1\right)}^{99}$

#### Explanation:

According to the chain rule:

$\frac{d}{\mathrm{dx}} \left[{\left({t}^{3} + 1\right)}^{100}\right] = 100 {\left({t}^{3} + 1\right)}^{99} \cdot \frac{d}{\mathrm{dx}} \left[{t}^{3} + 1\right]$

$= 100 {\left({t}^{3} + 1\right)}^{99} \cdot 3 {t}^{2}$

$= 300 {t}^{2} {\left({t}^{3} + 1\right)}^{99}$