How do you use the Product Rule to find the derivative of 3x(2x-1)^3?

Aug 9, 2015

${y}^{'} = 3 {\left(2 x - 1\right)}^{2} \cdot \left(8 x - 1\right)$

Explanation:

To be able to use the product rule to differentiate this function, you need to write it as a product of two other functions, let's say $f \left(x\right)$ and $g \left(x\right)$

$y = f \left(x\right) \cdot g \left(x\right)$

This will allow you to use the formula

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot g \left(x\right) + f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)}$

$y = {\underbrace{3 x}}_{\textcolor{b l u e}{f \left(x\right)}} \cdot {\underbrace{{\left(2 x - 1\right)}^{3}}}_{\textcolor{b l u e}{g \left(x\right)}}$

This means that you can write

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(3 x\right)\right] \cdot {\left(2 x - 1\right)}^{3} + 3 x \cdot \frac{d}{\mathrm{dx}} {\left(2 x - 1\right)}^{3}$

To calculate $\frac{d}{\mathrm{dx}} {\left(2 x - 1\right)}^{3}$, you can use the chain rule for ${u}^{3}$, with $u = 2 x - 1$.

$\frac{d}{\mathrm{dx}} \left({u}^{3}\right) = \frac{d}{\mathrm{du}} \left({u}^{3}\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

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

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

Your target derivative will thus be

${y}^{'} = 3 \cdot {\left(2 x - 1\right)}^{3} + 3 x \cdot 6 {\left(2 x - 1\right)}^{2}$

${y}^{'} = 3 {\left(2 x - 1\right)}^{3} + 18 x {\left(2 x - 1\right)}^{2}$

This can be further simplified to

${y}^{'} = 3 {\left(2 x - 1\right)}^{2} \cdot \left(2 x - 1 + 6 x\right)$

${y}^{'} = \textcolor{g r e e n}{3 {\left(2 x - 1\right)}^{2} \cdot \left(8 x - 1\right)}$