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Homework Help: Higher Derivative Question

  1. Oct 22, 2006 #1
    Alright I decided I'd create a new topic just because the other one was getting fairly lengthy.

    I'm having trouble with the following "Higher Derivative" question

    It states, find y'' by implicit differentiation.

    x^4 + y^4 = a^4

    d/dx (x^4+y^4) = d/dx (a^4)
    4x^3 + 4y^3 dy/dx = 4a^3

    However, what is the next step from here, I thought perhaps cancelling out all the 4's and leaving it as:
    x^3 + y^3 dy/dx = a^3
    and bring the x^3 over so it's:
    a^3-x^3 = y^3 dy/dx
    dy/dx = a^3-x^3 / y^3

    However, I'm not positive on that because I was told from a friend that 4a^3 is a constant so it can be equal to 0, thus the dy/dx would be -x^3/y^3.

    Am I on the right track here or is there an easier way of solving this equation. Thanks a lot guys.
  2. jcsd
  3. Oct 22, 2006 #2
    If 'a' is a constant, then its derivative (with respect to any variable) is zero.
  4. Oct 22, 2006 #3


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    4a^3 can't just be written as 0. But as neutrino pointed out, the derivative of a^4 isn't 4a^3
  5. Oct 22, 2006 #4
    Ahhh I think I'm following you, so pretty much from the first step you could automatically assume that a^4 it is equal to 0 when you take the derivative, and disregard the 4a^3 because that's not possible since it is a constant, thus after solving the problem it would be dy/dx = -x^3/y^3, but that's for y', not y''. You need to take the derivative of that now if I'm not mistaken using the rule:

    f'g-fg' / g^2

    And sub in:

    -x^3(y^3) - y^3(-x^3)

    And take the derivatives.
    Last edited: Oct 22, 2006
  6. Oct 22, 2006 #5


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    Correct. In fact, I'm sure you'll notice that once you use the chain rule, you'll get y'' in terms of x, y, and y' All you need to do is substitute -x3/y3 for y'
  7. Oct 22, 2006 #6
    You're right, and you can also take the derivative x^3 + y^3 dy/dx = 0, using the product rule for the second term.
  8. Oct 22, 2006 #7


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    Notice that when you differentiated y^4 with respect to x you did not get "4y^3", you got "4y^3 y' ". You could say that the derivative of a^4 with respect to x is 4a^3 a' and then since a is a constant, a'= 0 so (a^4)'= 4a^3(0)= 0. Of course, it's simpler to argue that a^4 is itself a constant so (a^4)'= 0 immediately.

    As neutrino said, once you have x^2+ y^3 y'= 0 you can just continue using "implicit differentiation": 2x+ 3y^2 (y')^2+ y^3 y"= 0
  9. Oct 22, 2006 #8
    Oh wow, alright thanks a lot for the advice all of you. HallsofIvy clarified that constant business for me perfectly. I understand that a whole lot better now. Except one correction in your description, it's x^3+y^3 y' = 0 believe as neutrino suggested, not x^2+ y^3 y'= 0
    Hence, it would be:
    3x^2 + 3y^2 (y')^2 + y^3 y'' = 0

    And then for y' you'd sub in -x^3/y^3 as Office Shredder suggested, and solve for y''?

    So it would be 3x^2 + 3y^2(-x^3/y^3)^2 + y^3 y'' = 0 I think anyways.
    Last edited: Oct 22, 2006
  10. Oct 22, 2006 #9
    Alright here's an update on what I did for this question anyways, because I assume you use the rule: f'g-fg'/g^2
    so from -x^3/y^3 to differentiate this expression I went:

    -3x^2(y^3) - (-x^3)(3y^2)y' / (y^3)^2
    = -3x^2(y^3) - (-x^3)(3y^2)(-x^3/y^3) / (y^3)^2
    Now how do I factor this expression.
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