# How do you use the chain rule to differentiate y=(2x^3+7)^6(2x+1)^8?

Nov 11, 2017

$36 {x}^{2} {\left(2 {x}^{3} + 7\right)}^{5} {\left(2 x + 1\right)}^{8} + 16 {\left(2 x + 1\right)}^{7} {\left(2 {x}^{3} + 7\right)}^{6}$

#### Explanation:

The first step here is to recognize that we are first going to use the product rule followed by the chain rule when we take the derivative:

$f ' \left(x\right) g \left(x\right) + g ' \left(x\right) f \left(x\right)$

It doesn't matter who is $f \left(x\right)$ or who is $g \left(x\right)$

I'll let $f \left(x\right)$ be the first function ${\left(2 {x}^{3} + 7\right)}^{6}$

And $g \left(x\right)$ be the second function ${\left(2 x + 1\right)}^{8}$

Following the product rule, lets do the first half:

$f ' \left(x\right) \times g \left(x\right)$

$f ' \left(x\right) \times g \left(x\right) = 6 {\left(2 {x}^{3} + 7\right)}^{5} \left(6 {x}^{2}\right) \times {\left(2 x + 1\right)}^{8}$

The $6 {x}^{2}$ came from the chain rule, where you take the derivative of the outside and then multiply it by derivative of the inside. Notice that the derivative of the inside function $\left(2 {x}^{3} + 7\right)$ is $6 {x}^{2}$ Our $g \left(x\right)$ remains untouched. We don't need the $\times$ sign I left it there to illustrate the point.

Now lets do the other half of product rule:

$g ' \left(x\right) f \left(x\right)$

$g ' \left(x\right) f \left(x\right) = 8 {\left(2 x + 1\right)}^{7} \left(2\right) \times {\left(2 {x}^{3} + 7\right)}^{6}$

The derivative of the inside function $\left(2 x + 1\right)$ is just $2$

Now just put them to together:

$6 {\left(2 {x}^{3} + 7\right)}^{5} \left(6 {x}^{2}\right) {\left(2 x + 1\right)}^{8} + 8 {\left(2 x + 1\right)}^{7} \left(2\right) {\left(2 {x}^{3} + 7\right)}^{6}$

You can simplify it:

$36 {x}^{2} {\left(2 {x}^{3} + 7\right)}^{5} {\left(2 x + 1\right)}^{8} + 16 {\left(2 x + 1\right)}^{7} {\left(2 {x}^{3} + 7\right)}^{6}$