Question

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

Answer 1

`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`

Answer 2

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.