Question

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

Answer 1

`36x^2(2x^3+7)^5(2x+1)^8+16(2x+1)^7(2x^3+7)^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'(x)g(x)+g'(x)f(x)`

It doesn't matter who is `f(x)` or who is `g(x)`

I'll let `f(x)` be the first function `(2x^3+7)^6`

And `g(x)` be the second function `(2x+1)^8`

Following the product rule, lets do the first half:

`f'(x) times g(x)`

`f'(x) times g(x)=6(2x^3+7)^5(6x^2)xx(2x+1)^8`

The `6x^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 `(2x^3+7)` is `6x^2` Our `g(x)` remains untouched. We don't need the `xx` sign I left it there to illustrate the point.

Now lets do the other half of product rule:

`g'(x)f(x)`

`g'(x)f(x)=8(2x+1)^7(2)xx(2x^3+7)^6`

The derivative of the inside function `(2x+1)` is just `2`

Now just put them to together:

`6(2x^3+7)^5(6x^2)(2x+1)^8+8(2x+1)^7(2)(2x^3+7)^6`

You can simplify it:

`36x^2(2x^3+7)^5(2x+1)^8+16(2x+1)^7(2x^3+7)^6`