# How do you find the derivative of (x^3+1)^2?

Jan 23, 2016

$\frac{d}{\mathrm{dx}} \left[{\left({x}^{3} + 1\right)}^{2}\right] = 6 {x}^{2} \left({x}^{3} + 1\right)$

#### Explanation:

Use the chain rule, which states that $\frac{d}{\mathrm{dx}} \left[{u}^{2}\right] = u ' \cdot 2 u$. Here, we know that $u = {x}^{3} + 1$, so

$\frac{d}{\mathrm{dx}} \left[{\left({x}^{3} + 1\right)}^{2}\right] = \frac{d}{\mathrm{dx}} \left[{x}^{3} + 1\right] \cdot 2 \left({x}^{3} + 1\right)$

$\frac{d}{\mathrm{dx}} \left[{\left({x}^{3} + 1\right)}^{2}\right] = 3 {x}^{2} \cdot 2 \left({x}^{3} + 1\right)$

$\frac{d}{\mathrm{dx}} \left[{\left({x}^{3} + 1\right)}^{2}\right] = 6 {x}^{2} \left({x}^{3} + 1\right)$