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Implicit differentiation - major confusion

  1. Apr 3, 2012 #1
    Hello. I know how to do implicit differentiation taught in calculus 1, but I'm confused by something regarding it.


    Take the example:

    y3+y2-5y-x2=4


    If we do implicit differentation we get:

    3y2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0

    dy/dx=2x/(3y2+2y-5)


    Now, it makes sense how to compute it, but I have troubles really understanding why it works. For example, if I graph the original function I find it fails the vertical line test (for example at -2) so I know it's not a function. But I only really understand what it means for a function to be differentiable. I guess what I mean is, when I think of something being differentiable, I think of a continuous function. What exactly is the original equation? How can I think of it as being continuous or differentiable or any such things? Can I use epsilon and delta definitions like I would normally to determine things like limits and continuity and differentiability?


    I hope this question makes sense. It's just, I know how to do the implicit differentiation, I just have difficulty thinking of taking the derivative of anything other than an actual function, if that makes sense.
     
  2. jcsd
  3. Apr 3, 2012 #2
    The other doubt I have here: if the original implicit equation is not actually a function, why do we assume y is based on x, which is the reason for the implicit differentiation using chain rule in the first place?
     
  4. Apr 3, 2012 #3

    Bacle2

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    If I understand you well, you do need conditions to define an implicit derivative; this has to see with the inverse function theorem and the implicit function theorems.
     
  5. Apr 3, 2012 #4
    To elaborate on what Bacle2 said, if you have some curve that's not a function, for any point (x,y) if you zoom in a lot it may pass the vertical line test for points in the immediate vicinity. In other words, the curve is not a function, but locally it looks like a function. So we can make it into a function in a small neighborhood surrounding a given point, and then we can take the derivative of that local function at that point. That's what we're really doing in implicit differentiation.
     
  6. Apr 3, 2012 #5

    Bacle2

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    One way of seeing why f'(x)=0 may give you problems, to follow up on Lugita15, is, for the case of n=2, that the problems may happen when f'=0 , since you may have a local max/min at a point, (so that, by definition, it will necessarily increase/decrease)which means that the function will take the same value y at least twice, and the inverse function will not exist.

    Note, tho, that the condition f'(x)≠0 is not necessary for a global inverse to exist, e.g.,
    f(x)=x1/3, which is globally-invertible, but f'(1/3)=0. This is because 0 is not an extreme value. Note also, in the case of complex functions, that for f(z)= ez, we have f'(z)≠ 0, but f does not have a global inverse.
     
  7. Apr 3, 2012 #6

    chiro

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    The reason you can do dy/dx(y^3) = 3y^2dy/dx is simply because of the chain rule. No other understanding is needed: if the assumptions are met for the chain rule, then all is well. The rest is just algebra.

    If you want to understand the chain rule in its generality then read a book on multivariable calculus. For specific examples consider a function y = u(v(x)) and see how the derivative dy/dx changes with u'(v(x)) and v'(x) by plotting a few sample functions. If you want to know this result mathematically have a look at the proof of the chain rule.
     
  8. Apr 3, 2012 #7

    Bacle2

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    The OP was referring, the way I saw it, to the fact there may be problems when the function is multivalued, so that you must select a "branch". Otherwise, how do you define f(x+h) and f(x) in the limit, for x multivalued? Which value of f(x+h),f(x)
    respectively , do you select?
     
  9. Apr 3, 2012 #8

    chiro

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    In this case, you already have the values (or should have) of both the y value and the x-value.

    consider y^2 + x^2 = 1 in which dy/dx = -x/y which is a multi-valued function. We have both y and x for this function which gives us the complete information for evaluating the derivative despite the fact that we have two-branches for this circle.

    The derivative still gives a valid answer in terms of how y changes with x even if y^2 + x^2 = 1 is not a valid function unless its restricted.

    All the dy/dx represents is how y changes with x in an increasing x given a current y and x and if you do an exercise where you start at the north pole and the south pole and calculate new approximate y values by using y(new) = yold + dy/dx x Δx for some appropriate Δx. You'll find that the produced values mimic the behaviour of the implicit representation as you would expect: the only difference is that you will have different y values for each 'branch'.

    Again, as long as the chain rule assumptions are satisfied for its application in this kind of circumstance (as well as any other derivative assumptions) then it doesn't matter if the function is multivalued or not.
     
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