# How do you differentiate y=(6x^2 + 2x)^3?

Mar 7, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 \left(12 x + 2\right) {\left(6 {x}^{2} + 2 x\right)}^{2}$

#### Explanation:

Don't bother expanding, just use the chain rule. Let $u = 6 {x}^{2} + 2 x$. Then $\mathrm{du} = 12 x + 2$. This also means that $y = {u}^{3}$., or $y = 3 {u}^{2}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(12 x + 2\right) 3 {\left(6 {x}^{2} + 2 x\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 \left(12 x + 2\right) {\left(6 {x}^{2} + 2 x\right)}^{2}$

Hopefully this helps!

Mar 7, 2018

$3 {\left(6 {x}^{2} + 2 x\right)}^{2} \left(12 x + 2\right)$

#### Explanation:

Applying the chain rule we arrive at the consensus that $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$
we identify the variables within our problem and define that $f \left(x\right) = {x}^{3}$ and
$f ' \left(x\right) = 3 {\left(x\right)}^{2}$
$g \left(x\right) = 6 {x}^{2} + 2 x$
combining the functions together with recieve the original function
$g ' \left(x\right) = 12 x + 2$ using the power rule $P {x}^{P - 1}$
from this point we plug out g(x) and f(x) functions into our rules of derivatives.