Question

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

Answer 1

`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`

Answer 2

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`