# How do you find the derivative of (4x^3+x^2)/2?

Apr 20, 2016

Use a little algebra and the power rule to find $\frac{d}{\mathrm{dx}} \frac{4 {x}^{3} + {x}^{2}}{2} = 6 {x}^{2} + x$.

#### Explanation:

Start by splitting this up into two fractions, like so:
$\frac{4 {x}^{3} + {x}^{2}}{2} = \frac{4 {x}^{3}}{2} + {x}^{2} / 2 = 2 {x}^{3} + {x}^{2} / 2$

Now, onto finding the derivative. The sum rule says we can break $\frac{d}{\mathrm{dx}} \left(2 {x}^{3} + {x}^{2} / 2\right)$ into $\frac{d}{\mathrm{dx}} \left(2 {x}^{3}\right) + \frac{d}{\mathrm{dx}} \left({x}^{2} / 2\right)$; in other words, we can take the derivative of a larger function piece by piece. We will evaluate both of these using the power rule:
$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$

Beginning with $\frac{d}{\mathrm{dx}} \left(2 {x}^{3}\right)$:
$\frac{d}{\mathrm{dx}} \left(2 {x}^{3}\right) = 3 \cdot 2 {x}^{3 - 1} = 6 {x}^{2}$

For $\frac{d}{\mathrm{dx}} \left({x}^{2} / 2\right)$, we have:
$\frac{d}{\mathrm{dx}} \left({x}^{2} / 2\right) = 2 \cdot {x}^{2 - 1} / 2 = {x}^{1} = x$

Putting these results back together yields:
$\frac{d}{\mathrm{dx}} \frac{4 {x}^{3} + {x}^{2}}{2} = 6 {x}^{2} + x$