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Differentiate the function

  1. May 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Differentiate:


    2. Relevant equations
    y=∛((7-3x^2)^2)

    3. The attempt at a solution
    y^'(x)=(4x(3x^2-7))/(∛((7-3x^2 )^2))^2 )
     
  2. jcsd
  3. May 15, 2013 #2

    Mark44

    Staff: Mentor

    It's much more convenient to write your function as y = (7 - 3x2)2/3. When you're differentiating, it's almost always better to rewrite expressions with radicals using exponents. Your answer might be correct, but if so, it needs to be simplified.
     
  4. May 16, 2013 #3
    I really need to know if it is correct, could you please let me know if it is?
     
  5. May 16, 2013 #4

    tiny-tim

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    (try using the X2 button just above the Reply box :wink:)
    that looks correct, but it needs simplifying :redface:

    (also, it's a ridiculous way of writing it, and you'll lose marks in the exam if you do that)
     
  6. May 16, 2013 #5
    Thank you for you reply. So what would be the correct way to write it so I do not lose marks for it?
     
  7. May 16, 2013 #6

    tiny-tim

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    first, simplify it, as Mark44 :smile: suggested
     
  8. May 16, 2013 #7
    Is this correct: y = (((7-3x^2)^2))^ (1/3)
    dy/dx = (1/3) (((7-3x^2)^2))^ (-2/3) (2(7-3x^2)(-6x))
    = (24/9)(7x - 3x^3) / ∛((7-3x^2)^4)
     
  9. May 16, 2013 #8

    tiny-tim

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    i'm sorry, but can't you see how ridiculous that is? :redface:

    why not just write y = (7-3x2)2/3 ? :confused:
     
  10. May 16, 2013 #9
    ok so is this a correct final answer now: y'(x)=4x(3x^2-7) ??
     
  11. May 16, 2013 #10
    The solution is

    12x(3x^(2)-7) isn't it?
     
  12. May 16, 2013 #11
    Sorry never mind thought you said derivative not differentiate!
     
  13. May 16, 2013 #12

    tiny-tim

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    how did you get that? :confused:
     
  14. May 16, 2013 #13

    Mark44

    Staff: Mentor

    No.
    You get the derivative when you differentiate a function. It seems that you didn't know this.
     
  15. May 16, 2013 #14
    I am really confused now, what is the correct answer then?
     
  16. May 16, 2013 #15

    Mark44

    Staff: Mentor

    This was the advice I gave in post #2.
    What do you get if you differentiate the function that I wrote? You need to use the chain rule (correctly).
     
  17. May 16, 2013 #16
    Differentiating the function you wrote would get: y'(x)=12x^3-28x
    Is this correct? Please help
     
  18. May 16, 2013 #17

    tiny-tim

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    no

    show us, step-by-step, how you got that
     
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