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Homework Help: Implicit Differentiation Question

  1. Oct 13, 2016 #1
    << Mentor Note -- thread moved from the technical math forums at OP request, so no Homework Help Template is shown >>

    x2y + xy2 = 6

    I know we use the chain rule from here, so wouldn't that be:

    (d/dx)(x2y + xy2) = (d/dx)(6)

    so using the chain rule of g'(x)f'(g(x) and the d/dx canceling out on the left side ...

    2x + (d/dy)(4y2) = 0

    Subtract 2x to the other side and cancel out the 4y2 and I got -x/2y2, which isn't right at all. :/
    Last edited by a moderator: Oct 13, 2016
  2. jcsd
  3. Oct 13, 2016 #2
    ...And I just saw in the rules I wasn't supposed to post this here. My bad, is there anyway to move it?
  4. Oct 13, 2016 #3


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    First thing: I assume, that ##y## isn't dependent on ##x##. You should mention this. Next, what happens to a factor when you differentiate?
  5. Oct 13, 2016 #4
    It says, "use implicit differentiation to find dy/dx," so sure?

    It becomes a derivative of the original?
  6. Oct 13, 2016 #5


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    Where did it say this?
    Anyway, if ##y=y(x)## then you have to use the product rule and the chain rule.

    Edit: I assume you meant the theorem of implicit functions.
  7. Oct 13, 2016 #6
    That's all the instructions said, although maybe the pages before had more information (I tired to read them but they didn't make any sense so I came here). To be frank I don't know why saying y isn't dependent on x is important or even quite what that means.

    The explanatory blue box doesn't mention theorems but it does say this:

    Implicit Differentiation
    1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x.
    2. Collect the terms with dy/dx on one side of the equation and solve for dy/dx.

    So I guess it's saying y IS dependent on x? Or they have a relationship? Which means it's a function, if that means anything?

    I don't know, this is really hard!
  8. Oct 13, 2016 #7
    I'll try it again with the product rule...
  9. Oct 13, 2016 #8

    Ray Vickson

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    You start with an equation involving both ##x## and ##y##, so for any given value of ##x## you can solve the equation to get one or more values of ##y##. In your case, when you give ##x## a value such as ##x = 1## or ##x = -7## (or whatever) you will get a quadratic equation in ##y## that has two roots that you can find by the elementary methods you studied years ago.

    So, indeed: we are interested in the case where y depends on x---very much so! The only problem is that although y depends on x we do not have an actual formula that tells us exactly what that dependence is. That is, while we know that we will have ##y = g(x)## for some function ##g## (that is, the function ##g(x)## exists), we don't have a formula for ##g(x)##. The question is asking you how you would find the derivative ##dy/dx= dg(x)/dx## without actually knowing the formula for ##g## in the relationship ##y = g(x)##.
  10. Oct 13, 2016 #9
    Okay, just to make sure I'm understanding:

    x + y = 6
    y = 6 - x
    So if x = 2, 4, 5, that would make y = 4, 2, 1, and so on.

    Hmm, okay. I don't really see it?

    Using the product rule I got:

    (2xy + x2) + (y2 + 2y) = 0

    The parentheses might be wrong, and I might've taken the derivative of 6 too soon, but I don't see where to go from here regardless. Surely it's not quadratic or going to be if there's no longer anything squared? Also... how do I chain 2xy?

    What you're saying makes sense but I don't see it in the problem. :eek: We have y = a function but we don't know what the function looks like (it's formula).

    On a side note, dy/dx = dg(x)/dx means
    the derivative of y in respect to the derivative of x = the derivative of function g(x) in respect to the same derivative of x

    ...but what does that MEAN?
  11. Oct 15, 2016 #10


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    That's incorrect.

    Looking at the derivative of your first term:

    The derivative (w.r.t. x) of x2y is:

    (the derivative of x2 ) times ( y )
    (x2 ) times ( the derivative if y )

    You have the first part right, but in the second part, you're missing the derivative of y .
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