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Homework Help: Implicit Differentiation

  1. Jun 22, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]x^{3}y + y^{3}x = 30[/itex]

    2. Relevant equations


    3. The attempt at a solution

    My understanding is that I'm supposed to take the derivative of each side and try to solve for y'.

    But, I don't know how to treat the y when I am differentiating it, to a power or whatever.

    For example, here I did

    [itex]Dx(x^{3}y) + Dx(y^{3}x) = Dx(30)[/itex]

    When differentiating the first term, using the product rule, I have to get Dx(y). Simply saying that it's 1 doesn't seem to work, so I just leave it as y'. Using the product rule on that first term on the left side gives me:

    y' + 3(x^2)y.

    Added to the derivative of (y^3)(x). But that's the derivative I'm having trouble with.

    Can anyone walk me through this? How do I treat y when differentiating a term with it in it? I don't understand.
  2. jcsd
  3. Jun 22, 2011 #2


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    Homework Helper

    think of y as a function of x, y = y(x)

    then think of the power as a function of y, g(y) = y^3

    then you have
    [tex] y^3 = (y(x))^3 = g(y(x))[/tex]

    now use the chain rule
    [tex] \frac{d}{dx}y^3 = \frac{d}{dx}g(y(x)) = g'(y(x))y'(x)[/tex]
    Last edited: Jun 23, 2011
  4. Jun 23, 2011 #3
    Will simply deriving it normally and then multiplying the result by y prime always give the same result as the chain rule?
  5. Jun 23, 2011 #4
    Alright, you know that feeling after something very simple clicks, and you kick yourself for not grasping it right off the bat, that's what I'm feeling now. I guess my textbook and I are on two different wavelengths...


    First term differentiated: 3yx^2+y'x^3
    Second term: 3xy'y^2 + y^3
    derivative of 30 is zero..

    So, 3yx^2 + y'x^3 + 3xy'y^2 + y^3 = 0

    Since I'm solving for y', get them all to once side..

    3yx^2 + y^3 = -y'x^3 - 3xy'y^2

    Factor out y' on right side.. : y'(-x^3 - 3xy^2), then divide for a final answer of:

    (3yx^2+y^3) / (-x^3 - 3xy^2) = y'

    Can someone confirm I'm on the right track?
  6. Jun 23, 2011 #5


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    Homework Helper

    look good to me
    [tex]y' = -\frac{y}{x} \frac{(3x^2+y^2} {x^2 + 3y^2} [/tex]
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