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\[f(x)=x ^{2}(x-2)^{4}\]

Use product rule,
y' = (x^2)(4)(x-2)^3+(x-2)^4 (2x)
\[4x^{2}(x-2)^{3}+2x(x-2)^{4}\]

yea product rule

you get 2x(x-2)^4+x^2(4(x-2)^3)

then apply the chain rule which i do not understand

Noneed to use chain rule here.

the answer in the back of the book says

\[x ^{2}[4(x-2)^{3}(1)] + (x-2)^{4}(2x) = 2x(x-2)^{3}(3x-2)\]

i dont understand hgow to get to 2x(x-2)^3(3x-2)

easy

After you do this about a 1,000 times you will be able to do it in your head

Any questions?
also if you do it in the order I do it you wont mess up using Quotient rule

sorry let me read what you wrote

had stepped away

this is the fool proof way of doing derivatives when you first start out at least it was for me

ok let me try to reword this

x^2(x+2)^4

(2x)((x+2)^4) + 4x^2(x-2)^3

ah so i then take (2x)((x+2)^4) + 4x^2(x-2)^3 and split it like [2x(x+2)^4 +4x^2](x-3)^3 ?

and do the product rule again?

no!

f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2)
That is it you can simplify it but that is the derivative

if you wanted to take the second derivative you would use product rule again.

ok so i need to simplify f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2)

so i factor out (x-2)^3

you can also factor out 2x

2x(x-2)^(3)(x - 2 + 2x)

so i would end up with 2x(x-2)^3(x-2+2x)

2x(3x-2)(x-2)^(3)

2x^(1) and x^(1) can be added together

you have been so much help

i had the right answer the whole time just didnt see the factor

*patronizing

yeah math can be a hassel sometimes lol

no i understand now

thanks a lot

np take it easy