1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    fresh_42

    Staff: Mentor

    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

    fresh_42

    Staff: Mentor

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

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

    You have the first part right, but in the second part, you're missing the derivative of y .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Implicit Differentiation Question
Loading...