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

Nov 2, 2015

Use chain rule / Expand

#### Explanation:

Method 1
$y = {\left(3 x + 1\right)}^{2}$
Assume the function in the bracket as some other variable say t.
$t = 3 x + 1$
$\frac{\mathrm{dt}}{\mathrm{dx}} = 3$
$y = {t}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$
$= 2 t \cdot 3$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 \left(3 x + 1\right)$

Method 2
If you can expand everything and find the derivative

$y = {\left(3 x + 1\right)}^{2}$
$y = 9 {x}^{2} + 6 x + 1$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 18 x + 6$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 \left(3 x + 1\right)$