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

Question

5366 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-21T04:22:00

#h'(x)=-24x^2+4x+15#

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

#u=3x-2x^2#
#u'=3-4x#

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

Substitute

#h'(x)=(3x-2x^2)(4)+(3-4x)(5+4x)#

Simplify

#h'(x)=(12x-8x^2)+15+12x-20x-16x^2#

Combine like terms

#h'(x)=4x-24x^2+15#

#h'(x)=-24x^2+4x+15#

Click here for another example