What is the derivative of #f(x) = x(2-x)^2#?

2 Answers

Answer:

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

Explanation:

The derivative using the product rule is

#(df)/dx=x'*(2-x)^2+x*[(2-x)^2]'=(2-x)^2+2*x*(-1)*(2-x)=(2-x)^2-2x*(2-x)= =3x^2-8x+4#

Expanding on the above
The Product Rule for Differentiation says that
if #color(white)("XX")g(x)=color(red)(a) * color(blue)(b)#
then #color(white)("XX")(d g(x))/(dx) = (dcolor(red)(a))/(dx) * color(blue)(b)+color(red)(a)*(dcolor(blue)(b))/dx#

For #f(x)=x(2-x)^2# we can treat #color(red)(x)# as #color(red)(a)a# and #color(blue)((2-x)^2)# as #color(blue)(b)#

So #color(green)((d f(x))/(dx) = dcolor(red)(x)/(dx) * color(blue)((2-x)^2) + color(red)(x) * (d(color(blue)((2-x)^2))/(dx))#

#dx/dx=1# so that part is easy

but #(d(2-x)^2)/(dx)=?# is a little more challenging.

Noting that #(2-x)^2=(color(orange)(2-x))(color(brown)(2-x))#
we can apply the Product Rule again to get
#color(white)("XXX")(d(2-x)^2)/(dx)=(d(color(orange)(2-x)))/dx * (color(brown)(2-x)) +(color(orange)(2-x)) * (d(color(brown)(2-x)))/dx#

Since #(d(2-x))/dx=-1#
we have
#color(white)("XXX")(d(2-x)^2)/(dx)= 2xx(-1)(2-x) = 2x^2-4#

and our original equation becomes
#color(green)((d f(x))/(dx) = (1) * (2-x)^2 + x * (2x-4))#
#color(white)("XXX")=(4-4x+x^2) +(2x^3-4x)#
#color(white)("XXX")=3x^2-8x+4#

Feb 17, 2016

Konstantios's answer provides some important concepts
but for this particular problem there may be a simpler method.

If we expand #x(2-x)^2#
we get
#color(white)("XXX")f(x)=x^3-4x^2+4x#

then using Exponent Reduction Rule for polynomial derivatives
#color(white)("XXX")f'(x)=3x^2-8x+4#