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

Question

Charles Anctil · Accepted Answer

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#

William Abdalla · Answer

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

The derivative using the product rule is

#(df)/dx=x'(2-x)^2+x[(2-x)^2]'=(2-x)^2+2x(-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#

Adam Adkins · Answer

undefined To find the derivative of the function $ f(x) = x(2-x)^2 $, you can use the product rule and the chain rule. Applying these rules:

$$ f'(x) = (1)(2-x)^2 + x \cdot 2(2-x)(-1) $$
$$ f'(x) = (2-x)^2 - 2x(2-x) $$
$$ f'(x) = (2-x)^2 - 4x(2-x) $$
$$ f'(x) = (2-x)^2 - 4x(2) + 4x^2 $$
$$ f'(x) = (2-x)^2 - 8x + 4x^2 $$
$$ f'(x) = 4x^2 - 4x + 4 - 8x + 4x^2 $$
$$ f'(x) = 8x^2 - 12x + 4 $$