How do you use the product rule to differentiate # y= 4x(2x+3)^2#?

2 Answers
Mar 4, 2016

Answer:

#dy/dx = 48x^2 + 96x + 36#

Explanation:

Using the#color(blue)" Product rule "#

If f(x) = g(x).h(x) then f'(x) = g(x)h'(x)+h(x)g'(x)

now g(x) = 4x #rArr g'(x) = 4 #

and #h(x)=(2x+3)^2 rArr h'(x) = 2(2x+3) d/dx(2x+3) #
Replace these results into f'(x)

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

#=16x(2x+3) +4(2x+3)^2=32x^2+48x+16x^2+48x+36#

#=48x^2 +96x + 36 #

Mar 4, 2016

Answer:

Let # f=4x, g=(2x+3)^2->f'=4,g'=2(2x+3)*2#
#y'=fg'+gf'=16x(2x+3)+4(2x+3)^2=32x^2+48x+16x^2+48x+36 = 48x^2+96x+36#

Explanation:

Separate the products into f and g then find each of the derivatives then plug it in to the product rule and simplify.