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Implicit differentiation

  1. Jul 18, 2011 #1

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 18, 2011 #2

    hunt_mat

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    There is only one way to treat this and that is via the chain rule, you know that [itex]y=y(x)[/itex] and so:
    [tex]
    \frac{d}{dx}=\frac{dy}{dx}\frac{d}{dy}
    [/tex]
    On a related note, you were wrong with your analysis because [itex]y=\pm\sqrt{1-x^{2}}[/itex].
     
    Last edited: Jul 18, 2011
  4. Jul 18, 2011 #3

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  5. Jul 18, 2011 #4
    No. Implicit means implied, but not plainly expressed. Implicit differentiation isn't going to be a function like y' = x, you're going to keep a y variable, otherwise it's explicit.

    Think about this: If I ask for dy/dx, I'm asking for derivative of function y with respect to variable x. I can't find the dy/dx of 2y, because there is no variable x in 2y! That's why we implicitly differentiate.

    First, let me give you a hand:

    [itex]x^2+y^2=1[/itex]

    First, find the dy/dx of [itex]x^2[/itex]. This has a variable of x, so it's just 2x.

    Next, find dy/dx of [itex]y^2[/itex]. 2y, right. Well, we have no x variable. So, include y' in this derivative. In other words, 2y times y', or 2yy'. Whenever the variable we are deriving in respect to is absent, we must multiply by y'. (that's why we need to multiply the entire derivative by dr/dt in your related rates problem ;) )

    Lastly, find dy/dx of 1. You know that the derivative of a constant is always 0.

    So now,

    [itex]2x + 2yy' = 0[/itex]

    Do you understand how I came to that result?

    Now, purely algebriacly, I want you to solve the equation for y'. Treat it as you would any variable.
     
  6. Jul 18, 2011 #5

    I like Serena

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    I'll just add another reply, which may help.

    Your differentiation of y2 should be:

    [tex]\frac d {dx}(y^2) = 2y \frac {dy} {dx}[/tex]

    This is the application of the chain rule.

    If you replace this in your equation, you should be able to solve the problem.
     
  7. Jul 18, 2011 #6

    Femme_physics

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    I guess my mistake was not differentiating y^2, and that's because I thought that the notation means to differentiate only x.


    But, apparently, you differentiate everything! But I don't get one thing


    In this


    [tex]\frac d {dx}(y^2) = 2y \frac {dy} {dx}[/tex]


    How come for the "y" expression you also have dy/dx, and for the x expression you don't?
     
  8. Jul 18, 2011 #7
    Because y and x variables are not equals here. dy/dx means derivative with respect to variable x. We include dy/dx with 2y because we have no variable x in that term.

    Just like if I asked dy/dx of 5t, 6q, 7u, or any other variable that was not x.

    Or if I asked dy/du of 5x, 6t, or any other variable that was not u.

    You have to include it when we have a "mismatched variable" to put it very informally. Chain rule shows this.
     
    Last edited: Jul 18, 2011
  9. Jul 18, 2011 #8

    Femme_physics

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  10. Jul 19, 2011 #9

    vela

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    You could look at it this way:[tex]\frac{d}{dx}(x^2) = 2x \frac{dx}{dx}[/tex]so it's the same as in the y case. But dx/dx=1, so you get[tex]\frac{d}{dx}(x^2) = 2x[/tex]
     
  11. Jul 19, 2011 #10

    Mark44

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  12. Jul 19, 2011 #11
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  13. Jul 19, 2011 #12

    Femme_physics

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    w00t!!! Thanks!! :biggrin: The idea of starting with simple problem is a lot better I see :approve:

    You guys rock for helping me! :smile:
     
  14. Jul 19, 2011 #13

    hunt_mat

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    It's why we're here.
     
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