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Implicit Differentiation

  1. Jul 26, 2007 #1
    Hi guys,

    I have a question on an implicit differentiation problem. I get two different answers depending on how I do it and the answers are different (not just looking, but different).

    The problem is [tex] \frac{x+3}{y}-4x-y^2=0 [\tex]. One option is to just differentiate as it stands and you get [tex] y'=\frac{y-4y^2}{x+3+2y^3} [\tex]. Another possibility is to initially multiply both sides of the equation by y to simplify it. If you do that the derivative you obtain is [tex] y'=\frac{1-4x}{4x-3y^2} [\tex]. These are two different derivatives but I'm not sure why this happens.

    Any thoughts would be appreciated (NOTE: I'm teaching the course and this came up).

    P.S. why aren't my equations tex-ing?

  2. jcsd
  3. Jul 26, 2007 #2
    Your eqns aren't texing because the end command is [ / tex] instead of your backslash.

    The point is that you define a function y implicitly. What makes you think you're allowed to multiply by y and obtain the same result? The statement you get surely is true (multiplying 0 by anything will yield 0, true), but it might simply define a different function y. See? You could multiply by anything, say xy, but the expression you get will not be equivalent to the thing you started with. When you multiply by 0 there's a loss of information.
  4. Jul 26, 2007 #3
    since they're not working, can you just type it out normally. i'm getting so confused as to what the problem is, lol.
  5. Jul 26, 2007 #4
    Here is the original problem:

    The problem is [tex] \frac{x+3}{y}-4x-y^2=0 [/tex]. One option is to just differentiate as it stands and you get [tex] y'=\frac{y-4y^2}{x+3+2y^3} [/tex]. Another possibility is to initially multiply both sides of the equation by y to simplify it. If you do that the derivative you obtain is [tex] y'=\frac{1-4y}{4x+3y^2} [/tex]. These are two different derivatives but I'm not sure why this happens.

    This is something subtle I think. If you take the original equation and multiply by y you get [tex]x+3-4xy-y^3=0[/tex]. You can solve for [tex]x+3[/tex] to get [tex]x+3 = 4xy+y^3[/tex]. Then plug this into the derivative obtained by straight implicit differentiation (i.e. [tex] y'=\frac{y-4y^2}{x+3+2y^3} [/tex]) and with a bit of algebra you get the other function.

    Now, I plotted both of these "derivatives" as functions of (x,y) and they are different functions. However, when we restrict their domain to that subset of the plane which represents the graph of the original function, they agree. That strikes me as interesting. How many functions could you have that all agreed on that same point set that you could arrive at through implicit differentiation? Probably a lot.

    Any thoughts or comments would be well appreciated,

    Last edited: Jul 27, 2007
  6. Jul 26, 2007 #5
    Did you read my post?
  7. Jul 26, 2007 #6
    could you show your steps for y' = 1-4x / 4x-3y^2

    this so far is your best solution but it's still not right. kind of close tho.
  8. Jul 26, 2007 #7
    why is that so? As long as y is a function of x, and y is non-zero, why is multiplying the equation by y leading to a different deriavative of y?
  9. Jul 26, 2007 #8
    Oops, that was a typo, I'll edit the post
  10. Jul 26, 2007 #9

    I didn't respond, because it doesn't matter. y=0 isn't in the domain, multiplying by y is find otherwise.

    Thanks though,

  11. Jul 27, 2007 #10
    Both the expressions for y' are correct. easy manipulation of y^3 will get u the equality.

    the second expression for y' is wrong
  12. Jul 27, 2007 #11
    RohanSingh: did you mean the minus sign in the bottom, sorry another typo. Nevertheless, noone seems particularly interested in the nature of the problem. The two derivatives are not, in general equal, they are only equal on limited portions of their domains, in particular the point set that represents the graph of the implicit function. That's interesting to me.

    So, for the record, I know both derivatives are correct (typos aside).


  13. Jul 27, 2007 #12
    If you multiply by y, your answer will be the same as long as you stick with your initial domain, which excludes y=0. So surely the two results must agree on the level sets of your first equation. But, that's what I was trying to say, if you multiply by anything you might get a new level set, which contains the old one, and new values for the derivative. Try multiplying by xy for example, and you get a new answer. That however isn't surprising, because the domain of y' as you calculate it implicitly really only are those x where y(x) is part of the level set implicitly defined. The algebraic expression you get may be extendable, but that concerns a new, different function then.

    Well, the two derivatives are not defined in general! My differentiating implicitly you assume that your always on the level set. As I mentioned above, there's no validity to the derivatives outside the domain of the original function. Do you know what I mean?
    Last edited: Jul 27, 2007
  14. Jul 27, 2007 #13
    It is easy to see that both are equal by dividing the top and bottom of the first expression by y, and then using the definition of y.
  15. Jul 27, 2007 #14
    Thanks, DeadWolfe, that's exactly what I'm saying: You need to use the definition of y, the level set.
  16. Jul 27, 2007 #15
    Thanks for the discussion guys. Nevertheless, it never occured to me that this happens. It doesn't happen for functions, just implicit functions. As derivatives they're not defined on anything but the limit set, but they are functions in their own right, and I still think its interesting. I just think there might be some interesting math in here. Differentaition of implicitly defined curves maps maps these curves into subsets of a function space. I haven't thought about it much, but it might be fun to see if the functions that agree on the level sets could form some sort of equivalence class and so you could look at some sort of quotient structure and see what comes out of it.

    I'm excited because this makes implicit functions interesting to me, I haven't ever seen much use in them so its nice to ponder something.

    Thanks again for your comments,

  17. Jul 27, 2007 #16
    you can definitely apply this technique for optimization.

    in Stewart's Calculus, he did an example that took so many damn steps, then he showed it through implicit differentiation and it was like 1/4 the agony of doing it the long way ... i was like wow.
    Last edited: Jul 27, 2007
  18. Jul 29, 2007 #17
    I don't have a copy of stewart's, could you outline the problem statement?


  19. Jul 29, 2007 #18
    A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can.

    You need the Volume & SA Formula.
  20. Aug 3, 2007 #19
    Thanks for the example, however I don't see why you'd need implicit differentiation. Set V=1 and solve for h in the formula for a volume of a cylinder and jam that into the formula for Surface area of a cylinder and differentiate it. It is minimized by a radius of (1/(2pi))^(1/3) and a height of (1/pi)(2pi)^(2/3). Seems pretty standard. I'm still looking for an example that would justify implicit differentiation.


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