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Trouble With Understanding Implicit Differentiation

  1. Jul 4, 2011 #1
    I don't think I fully understand implicit differentiation. I have read my textbook and watched many videos, and I think I will get an A on my test on this solely by memorizing the rules, but I would really like to understand this topic. From what I know, you are supposed to use implicit differentiation when the variable of the thing you are taking the derivative of disagrees with the denominator of d/dx and when they agree use explicit differentiation. I understand that there is a problem with taking the derivative of an expression when the variables disagree, but I don't understand why you can use the chain rule when the variables disagree but you cannot use the simple power rule (i.e. d/dx [yˆ2] = 2y(dy/dx) ≠ 2y). It seems arbitrary; why are you allowed to use the chain rule and not the simple power rule?

    Thanks
     
  2. jcsd
  3. Jul 4, 2011 #2

    lanedance

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    the difference is you are differentiating y = y(x), a fucntion of x, with respect to x, so you keep the dy/dx term

    if you were differentiating with respect to y then the power law applies
    [tex] \frac{d}{dy}y^2 = 2y [/tex]
     
  4. Jul 4, 2011 #3

    lanedance

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    and in fact you are using both the chain rule and the power rule
     
  5. Jul 4, 2011 #4
    ^I know, but why? Why do you only use the chain rule when the variables disagree?
     
  6. Jul 4, 2011 #5
    I think you actually use the chain rule when the variable agrees. It's just that the derivative of the inside of say X^3 is 1, (that is, the derivative with respect to x of x). So you have something like this:
    [tex]\frac{d}{dx}x^{3}=3x^{2}(\frac{d}{dx}x)[/tex]
    The latter part is just one, so we don't have to write it.

    Think of it like a composition:
    [tex]f(x) = x^{3}[/tex]
    [tex]g(x) = x[/tex]
    [tex](f \circ g)(x) = f(g(x)) = (x)^3[/tex]
     
    Last edited: Jul 4, 2011
  7. Jul 4, 2011 #6

    lanedance

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    yeah so just to add to quarkcharmers post

    consider the function f(x) = x

    then consider g(f) = f^2

    taking teh composition of functions we have
    [tex] g(f(x)) = (f(x))^2 = x^2[/tex]

    when we differntiate w.r.t. x
    [tex] \frac{d}{dx}g(f(x)) = \frac{dg(f(x))}{dx}g(f(x))\frac{df(x)}{dx} = g'(f(x))f'(x) = 2f(x).1 = 2x[/tex]

    so the key part isfor f(x) = x
    [tex] \frac{df(x)}{dx} = f'(x) = \frac{d(x)}{dx} = 1[/tex]
     
    Last edited: Jul 4, 2011
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