# How do you differentiate  y = ((2x - 1)^3)((x + 1)^3)?

##### 1 Answer
Sep 14, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\left(2 x - 1\right)}^{2} {\left(x + 1\right)}^{2} \left(4 x + 1\right)$

#### Explanation:

Product rule states if $y \left(x\right) = g \left(x\right) h \left(x\right)$

then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right) + \frac{\mathrm{dh}}{\mathrm{dx}} \times g \left(x\right)$

Hence as $y \left(x\right) = {\left(2 x - 1\right)}^{3} {\left(x + 1\right)}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 \times {\left(2 x - 1\right)}^{2} \times 2 \times {\left(x + 1\right)}^{3} + 3 \times {\left(x + 1\right)}^{2} \times 1 \times {\left(2 x - 1\right)}^{3}$

Here we have also used the concept of function of a function and used chain rule for it. As we have differentiated ${\left(2 x - 1\right)}^{3}$ and ${\left(x + 1\right)}^{3}$ w.r.t. $\left(2 x - 1\right)$ and $\left(x + 1\right)$, we must multiply by differential of $\left(2 x - 1\right)$ and $\left(x + 1\right)$ w.r.t $x$ i.e. by $2$ and $1$ respectively

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

= $3 {\left(2 x - 1\right)}^{2} {\left(x + 1\right)}^{2} \left[2 \left(x + 1\right) + 2 x - 1\right]$

= $3 {\left(2 x - 1\right)}^{2} {\left(x + 1\right)}^{2} \left[2 x + 2 + 2 x - 1\right]$

= $3 {\left(2 x - 1\right)}^{2} {\left(x + 1\right)}^{2} \left(4 x + 1\right)$