How do you differentiate #f(x)=4x(2x+3)^2 # using the product rule?

1 Answer
Jan 31, 2016

Answer:

#f'(x)=12(2x+3)(2x+1)#

Explanation:

The product rule states that for a function #f(x)=g(x)h(x)#, the derivative of the function is

#f'(x)=g'(x)h(x)+g(x)h'(x)#

Applying this to #f(x)=4x(2x+3)^2#, we see that

#f'(x)=color(red)(d/dx[4x])(2x+3)^2+4xcolor(blue)(d/dx[(2x+3)^2])#

We can find each of these derivatives separately and then plug them back in.

#color(red)(d/dx[4x]=4#

The following will require the chain rule.

#color(blue)(d/dx[(2x+3)^2])=2(2x+3)^1d/dx[2x+3]color(blue)(=4(2x+3)#

From these, we see that

#f'(x)=4(2x+3)^2+4x(4(2x+3))#

Factor out a #(2x+3)# term.

#f'(x)=(2x+3)(4(2x+3)+16x)#

Simplify.

#f'(x)=12(2x+3)(2x+1)#

This is also equal to

#f'(x)=48x^2+96x+36#