# How do you find (dy)/(dx) given y=x^3/3-1?

Apr 2, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2}$

#### Explanation:

We have:

$y = {x}^{3} / 3 - 1$ Apply $\frac{d}{\mathrm{dx}}$ on both sides.

$\implies \frac{d}{\mathrm{dx}} \left[y\right] = \frac{d}{\mathrm{dx}} \left[\frac{1}{3} {x}^{3} - 1\right]$

$\implies \frac{d}{\mathrm{dx}} \left[y\right] = \frac{d}{\mathrm{dx}} \left[\frac{1}{3} {x}^{3}\right] - \frac{d}{\mathrm{dx}} \left[1\right]$

A derivative of a constant is always 0.

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \cdot \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] - 0$

$\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$ if $n$ is a constant.

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \cdot 3 \cdot {x}^{3 - 1} - 0$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 1 \cdot {x}^{2} - 0$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2}$