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Find the derivative of the funtion. f(x)=x^2(x-2)^4

Mathematics
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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) \[4x^{2}(x-2)^{3}+2x(x-2)^{4}\]
Teach you one technique, y=(left)(right) y'=(copy left)(differentiate right)+(copy right)(differentiate left)

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Other answers:

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)
f(x)=x^2(x-2)^4 split these things up durp, g(x) = x^(2) g'(x) = 2x d(x) = (x-2)^4 USE CHAIN RULE!!! s(x) = x^4 s'(x) = 4x^(3) j(x) = x-2 j'(x) = 1 so d'(x) =4(x-2)^(3)*1 so Now product rule g(x) = x^(2) g'(x) = 2x d(x) = (x-2)^4 d'(x) =4(x-2)^(3)*1 2x((x-2)^4) + (4(x-2)^(3))x^(2)
easy
f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2) Best way to deal with derivatives when you are first learning them is to split them into different functions and derive them then put them all together
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
Remember Chain rule is just g(x) = x^4 g'(x) = 4x^(3) s(x) = x-2 s'(x) = 1 we throw away g(x) and sub s(x) into g'(x) and multiply by s(x) derivative so g'(s(x)) * s'(x)
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
f(x)=x^2(x-2)^4 f'(x) = 2x *(x-2)^4 + 4x^2 (x -2) ^3 = (x - 3) ^3 [ 2x(x-2) + 4x^2 ] = (x - 3) ^3 [ 6x^2 - 4x ]
(2x)((x+2)^4) + 4x^2(x-2)^3
g(x)d(x) Take derivative of g(x) take derivative of d(x) you have to use chain rule to do so g'(x)d(x) + g(x)d'(x) if you do it in this order it works for quient rule as well g(x)/d(x) (g'(x)(d(x)) - g(x)d'(x))/(d(x))^(2)
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)
its only asking for the first derivative so yeah. Is there anything you do not understand about the method I showed you? If you have to you can
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
because they are to the same power, remember they have to be to the same power to add them you can't add x^(2) + x^(1) or x^(3) + x Not to be partrionizing
*patronizing
yeah math can be a hassel sometimes lol
no i understand now
thanks a lot
np take it easy

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