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Implicit Differentiation: Differentiating in Terms of X

  1. Mar 29, 2015 #1

    Drakkith

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    I'm having some trouble with the terminology used in calculus.

    My book states: "Fortunately we don't need to solve an equation for Y in terms of X in order to find the derivative of Y. Instead we can use the method of implicit differentiation. This consists of differentiating both sides of the equation with respect to X and then solving the resulting equation for Y'."

    And: "In the examples and exercises of this section it is always assumed that the given equation determines Y implicitly as a differentiable function of X so that the method of implicit differentiation can be applied."

    I don't quite understand what they are telling me here. What does "differentiating with respect to X" and "Y as a differentiable function of X" mean?

    If it helps explain the above, here's an example equation from the section: X2+Y2 = 25
     
  2. jcsd
  3. Mar 29, 2015 #2

    Mark44

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    Differentiating with respect to x means applying the ##\frac d {dx}## operator. If we differentiate both sides of the equation wrt x, we get:
    ##\frac d {dx} (x^2 + y^2) = \frac d {dx} 25##
    ##\Rightarrow \frac d {dx} x^2 + \frac d {dx} y^2 = 0##
    ##\Rightarrow 2x + \frac d {dx} y^2 = 0##

    To differentiate y2 with respect to x, we need to use the chain rule, something you asked about in a previous thread.
    ## \frac d {dx} y^2 = \frac d {dy} y^2 \frac{dy}{dx} = 2y \frac {dy}{dx} = 2y * y'##
     
    Last edited by a moderator: Apr 2, 2015
  4. Mar 29, 2015 #3

    Drakkith

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    So "differentiating Y2 with respect to X" just means we are doing F'(x)?

    Thinking more about the second question, I'm guessing that "Y as a differentiable function of X" means that Y = F(x) where F(x) is a differentiable function?
     
  5. Mar 30, 2015 #4

    Mark44

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    No. You don't "do" F'(x).
    F'(x) is a thing (a noun). Differentiating is an action (a verb).

    F'(x) is what you get by differentiating F(x). IOW, ##\frac d {dx} F(x) = F'(x)##.
    Sort of, but you don't need F there. if y is a differentiable function of x, then dy/dx exists.
     
  6. Mar 30, 2015 #5

    Drakkith

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    Alright, thanks Mark.
     
  7. Apr 3, 2015 #6

    NascentOxygen

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    Taking your sample equation, there are two routes by which you can find ##\frac {dy} {dx}##

    Method 1: re-arrange the equation to isolate y on one side by itself,

    ##y\ =\ ±\sqrt{25\ -\ x^2}##

    and now differentiate both side with respect to x. (This is the method you most likely would have used.)

    Method 2: don't rearrange the equation; keep it as just as it is and differentiate both sides with respect to x. Mark44 has demonstrated Method 2, though hasn't finished it to the point of expressing dy/dx as a function of x only (because often this last step is not necessary). But perhaps as an exercise you could complete it, to show dy/dx as a function of x only. The final answer will be the same as for method 1. https://www.physicsforums.com/threa...rentiating-in-terms-of-x.805803/#post-5058053

    The reason your textbook implies that Method 2 is easier/shorter may not yet be apparent. But if your task were to determine the value of ##\frac {dy} {dx}## at a given point, say (3,-4), then Method 2 may turn out easier. For more complicated equations it generally will.
     
  8. Apr 3, 2015 #7

    micromass

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    If you're still having trouble, then perhaps replace ##y## with ##f(x)## and do what you usually do.

    So if you have ##x^2 + y^2 = 2##. Then you should see it as ##x^2 + f(x)^2 =2##. Then differentiate both sides to get ##2x + 2f(x)f^\prime(x) = 0##.

    By the way, if you're like me and often forget the quotient rule, then you can do it with implicit differentiation. So if ##f(x) = \frac{F(x)}{G(x)}##. Then rewrite it as ##f(x)G(x) = F(x)##. Differentiate both sides to get ##f^\prime(x)G(x) + f(x)G^\prime(x) = F^\prime(x)##. Isolate ##f^\prime(x)## to get

    [tex]f^\prime(x) = \frac{F^\prime(x) - f(x)G^\prime(x)}{G(x)}= \frac{F^\prime(x) - \frac{F(x)}{G(x)} G^\prime(x)}{G(x)} = \frac{ F^\prime(x)G(x) - F(x)G^\prime(x)}{G(x)^2}.[/tex]

    So you can rediscover the formula like this, or just apply it to a special case. Like if you have ##f(x) = \frac{x^2 + 1}{2x}##, then just differentiate both sides ##2xf(x) = x^2 + 1##.
     
  9. Apr 8, 2015 #8

    Drakkith

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    Thanks guys. After doing more of these problems it's a little clearer on what's going on.
     
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