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ChrisV

  • 2 years ago

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

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  1. ChrisV
    • 2 years ago
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    \[f(x)=x ^{2}(x-2)^{4}\]

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

  3. .Sam.
    • 2 years ago
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    Teach you one technique, y=(left)(right) y'=(copy left)(differentiate right)+(copy right)(differentiate left)

  4. ChrisV
    • 2 years ago
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    yea product rule

  5. ChrisV
    • 2 years ago
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    you get 2x(x-2)^4+x^2(4(x-2)^3)

  6. ChrisV
    • 2 years ago
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    then apply the chain rule which i do not understand

  7. .Sam.
    • 2 years ago
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    Noneed to use chain rule here.

  8. ChrisV
    • 2 years ago
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    the answer in the back of the book says

  9. ChrisV
    • 2 years ago
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    \[x ^{2}[4(x-2)^{3}(1)] + (x-2)^{4}(2x) = 2x(x-2)^{3}(3x-2)\]

  10. ChrisV
    • 2 years ago
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    i dont understand hgow to get to 2x(x-2)^3(3x-2)

  11. cuddlepony
    • 2 years ago
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    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)

  12. cuddlepony
    • 2 years ago
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    easy

  13. cuddlepony
    • 2 years ago
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    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

  14. cuddlepony
    • 2 years ago
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    After you do this about a 1,000 times you will be able to do it in your head

  15. cuddlepony
    • 2 years ago
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    Any questions? also if you do it in the order I do it you wont mess up using Quotient rule

  16. ChrisV
    • 2 years ago
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    sorry let me read what you wrote

  17. ChrisV
    • 2 years ago
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    had stepped away

  18. cuddlepony
    • 2 years ago
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    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)

  19. cuddlepony
    • 2 years ago
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    this is the fool proof way of doing derivatives when you first start out at least it was for me

  20. ChrisV
    • 2 years ago
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    ok let me try to reword this

  21. ChrisV
    • 2 years ago
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    x^2(x+2)^4

  22. Chlorophyll
    • 2 years ago
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    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 ]

  23. ChrisV
    • 2 years ago
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    (2x)((x+2)^4) + 4x^2(x-2)^3

  24. cuddlepony
    • 2 years ago
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    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)

  25. ChrisV
    • 2 years ago
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    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 ?

  26. ChrisV
    • 2 years ago
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    and do the product rule again?

  27. cuddlepony
    • 2 years ago
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    no!

  28. cuddlepony
    • 2 years ago
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    f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2) That is it you can simplify it but that is the derivative

  29. cuddlepony
    • 2 years ago
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    if you wanted to take the second derivative you would use product rule again.

  30. ChrisV
    • 2 years ago
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    ok so i need to simplify f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2)

  31. cuddlepony
    • 2 years ago
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    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

  32. ChrisV
    • 2 years ago
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    so i factor out (x-2)^3

  33. cuddlepony
    • 2 years ago
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    you can also factor out 2x

  34. cuddlepony
    • 2 years ago
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    2x(x-2)^(3)(x - 2 + 2x)

  35. ChrisV
    • 2 years ago
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    so i would end up with 2x(x-2)^3(x-2+2x)

  36. cuddlepony
    • 2 years ago
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    2x(3x-2)(x-2)^(3)

  37. cuddlepony
    • 2 years ago
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    2x^(1) and x^(1) can be added together

  38. ChrisV
    • 2 years ago
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    you have been so much help

  39. ChrisV
    • 2 years ago
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    i had the right answer the whole time just didnt see the factor

  40. cuddlepony
    • 2 years ago
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    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

  41. cuddlepony
    • 2 years ago
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    *patronizing

  42. cuddlepony
    • 2 years ago
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    yeah math can be a hassel sometimes lol

  43. ChrisV
    • 2 years ago
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    no i understand now

  44. ChrisV
    • 2 years ago
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    thanks a lot

  45. cuddlepony
    • 2 years ago
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    np take it easy

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