How do you differentiate $h (x) = (3 x - 2 x^{2}) (5 + 4 x)$ $h(x)=(3x-2x^2)(5+4x)$ using the product rule?

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Answer 1 · 2016-10-21T04:22:00

$h' (x) = - 24 x^{2} + 4 x + 15$ $h'(x)=-24x^2+4x+15$

$h' (x) = u v' + u' v$ $h'(x)=uv'+u'v$

$u = 3 x - 2 x^{2}$ $u=3x-2x^2$
$u' = 3 - 4 x$ $u'=3-4x$

$v = 5 + 4 x$ $v=5+4x$
$v' = 0 + 4 = 4$ $v'=0+4=4$

Substitute

$h' (x) = (3 x - 2 x^{2}) (4) + (3 - 4 x) (5 + 4 x)$ $h'(x)=(3x-2x^2)(4)+(3-4x)(5+4x)$

Simplify

$h' (x) = (12 x - 8 x^{2}) + 15 + 12 x - 20 x - 16 x^{2}$ $h'(x)=(12x-8x^2)+15+12x-20x-16x^2$

Combine like terms

$h' (x) = 4 x - 24 x^{2} + 15$ $h'(x)=4x-24x^2+15$

$h' (x) = - 24 x^{2} + 4 x + 15$ $h'(x)=-24x^2+4x+15$

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How do you differentiate h(x)=(3x−2x2)(5+4x)h(x)=(3x-2x^2)(5+4x) using the product rule?