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#