anonymous
  • anonymous
Find the derivative of the funtion. f(x)=x^2(x-2)^4
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[f(x)=x ^{2}(x-2)^{4}\]
.Sam.
  • .Sam.
Use product rule, y' = (x^2)(4)(x-2)^3+(x-2)^4 (2x) \[4x^{2}(x-2)^{3}+2x(x-2)^{4}\]
.Sam.
  • .Sam.
Teach you one technique, y=(left)(right) y'=(copy left)(differentiate right)+(copy right)(differentiate left)

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anonymous
  • anonymous
yea product rule
anonymous
  • anonymous
you get 2x(x-2)^4+x^2(4(x-2)^3)
anonymous
  • anonymous
then apply the chain rule which i do not understand
.Sam.
  • .Sam.
Noneed to use chain rule here.
anonymous
  • anonymous
the answer in the back of the book says
anonymous
  • anonymous
\[x ^{2}[4(x-2)^{3}(1)] + (x-2)^{4}(2x) = 2x(x-2)^{3}(3x-2)\]
anonymous
  • anonymous
i dont understand hgow to get to 2x(x-2)^3(3x-2)
anonymous
  • anonymous
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)
anonymous
  • anonymous
easy
anonymous
  • anonymous
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
anonymous
  • anonymous
After you do this about a 1,000 times you will be able to do it in your head
anonymous
  • anonymous
Any questions? also if you do it in the order I do it you wont mess up using Quotient rule
anonymous
  • anonymous
sorry let me read what you wrote
anonymous
  • anonymous
had stepped away
anonymous
  • anonymous
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)
anonymous
  • anonymous
this is the fool proof way of doing derivatives when you first start out at least it was for me
anonymous
  • anonymous
ok let me try to reword this
anonymous
  • anonymous
x^2(x+2)^4
anonymous
  • anonymous
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 ]
anonymous
  • anonymous
(2x)((x+2)^4) + 4x^2(x-2)^3
anonymous
  • anonymous
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)
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
and do the product rule again?
anonymous
  • anonymous
no!
anonymous
  • anonymous
f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2) That is it you can simplify it but that is the derivative
anonymous
  • anonymous
if you wanted to take the second derivative you would use product rule again.
anonymous
  • anonymous
ok so i need to simplify f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2)
anonymous
  • anonymous
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
anonymous
  • anonymous
so i factor out (x-2)^3
anonymous
  • anonymous
you can also factor out 2x
anonymous
  • anonymous
2x(x-2)^(3)(x - 2 + 2x)
anonymous
  • anonymous
so i would end up with 2x(x-2)^3(x-2+2x)
anonymous
  • anonymous
2x(3x-2)(x-2)^(3)
anonymous
  • anonymous
2x^(1) and x^(1) can be added together
anonymous
  • anonymous
you have been so much help
anonymous
  • anonymous
i had the right answer the whole time just didnt see the factor
anonymous
  • anonymous
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
anonymous
  • anonymous
*patronizing
anonymous
  • anonymous
yeah math can be a hassel sometimes lol
anonymous
  • anonymous
no i understand now
anonymous
  • anonymous
thanks a lot
anonymous
  • anonymous
np take it easy

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