How do you find the derivative of #(4x+1)^2(1-x)^3#?

2 Answers
Oct 5, 2016

Answer:

Use product and chain rule

Explanation:

#f'(x)=u'v+v'u#
Let #u=(4x+1)^2, v=(1-x)^3#
#u'=8(4x+1), v'=-3(1-x)^2#
#f'(x)=8(4x+1)(1-x)^3-3(1-x)^2(4x+1)^2#
#=(4x+1)(1-x)^2(8(1-x)-3(4x+1))#
#=(4x+1)(1-x)^2(5-20x)#

Oct 5, 2016

Answer:

#f'(x)=-3(4x+1)^2(1-x)^2+8(4x+1)(1-x)^3#

Explanation:

You have to use a combination of the Product Rule and Chain Rule .

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

#u=(4x+1)^2#
#u'=2(4x+1)*4 =># APPLY Chain Rule

#v=(1-x)^3#
#v'=3(1-x)^2*-1 =># APPLY Chain Rule

#f'(x)=(4x+1)^2 3(1-x)^2*-1+2(4x+1)*4(1-x)^3#

#f'(x)=-3(4x+1)^2(1-x)^2+8(4x+1)(1-x)^3#