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Flip dx/dy according to chain rule

  1. Jun 12, 2010 #1
    Hi

    I've just been reading something which is essentially how to work out what the deriviative of y=b^x is.

    Basically the explanation gets to the point which I understand and says

    [tex]\frac{dx}{dy} = \frac{1}{yln(b)}[/tex]

    It then says because of the chain rule you can simply flip this to get

    [tex]\frac{dy}{dx} = yln(b)[/tex]

    I have used the chain rule and thought I understood it but perhaps I don't because I don't see how the chain rule lets you do the 'flip' just described.

    Any comments greatly welcomed

    Kind regards
     
  2. jcsd
  3. Jun 12, 2010 #2

    tiny-tim

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    Hi Moogie! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    If y is a function of x, and x is a function of t, then of course the chain rule gives

    dy/dt = dy/dx dx/dt​

    Now put y = t … then we get

    1 = dy/dy = dy/dx dx/dy :wink:
     
  4. Jun 12, 2010 #3
    Hi

    Thanks for your reply but I don't understand how that answers my question.

    I think you are trying to show me that dy/dx * dx/dy = 1 but I don't understand why (my limitation, not your explanation).

    Could you try and explain it again from a different perspective? I didn't get the y=t thing. I can see how the 'symbols' make sense but I have no understanding or intuition of whats going on behind the symbols and thats more important to me.

    thank you
     
  5. Jun 12, 2010 #4

    Mark44

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    If the product of two expressions is 1, the expressions are reciprocals of each other.

    In tiny-tim's work, if y = t, then dy/dt = 1.
     
  6. Jun 13, 2010 #5
    Hi

    I understand that if the product of 2 expressions is 1 then they are reciprocals. I failed to understand where the chain rule came into it. How does the chain rule tell me that dy/dx * dx/dy = 1?

    I was hoping someone might be able to explain it to me from a different perspective (unless it is that simple that there isn't another perspective)
     
  7. Jun 13, 2010 #6

    tiny-tim

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    Hi Moogie! :smile:

    I don't think there is any different perspective

    Do you accept that dy/dt = dy/dx dx/dt comes from the chain rule?

    If so, the result does come from putting t = y.
     
  8. Jun 13, 2010 #7

    HallsofIvy

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    Then, what, exactly, do you understand as the "chain rule"?
     
  9. Jun 13, 2010 #8
    Hi Tiny Tim

    I do see that dy/dt = dy/dx dx/dt from the chain rule. Perhaps I am missing what it is you are doing when you set t = y

    thanks
     
  10. Jun 13, 2010 #9

    vela

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    Say you have a function y=f(x) and it has an inverse x=g(y). Then dy/dx=f'(x) and dx/dy=g'(y). Because f and g are inverses, you have x=g(f(x)). If you differentiate this, you get, using the chain rule,

    [tex]x=g(f(x)) \hspace{0.3in} \Rightarrow \hspace{0.3in} 1=g'(f(x))f'(x)=g'(y)f'(x)=\frac{dx}{dy}\frac{dy}{dx}[/tex]
     
  11. Jun 14, 2010 #10
    The explanation with inverses made more sense to me. But now I'm thinking perhaps I don't understand the chain rule after all.

    When you go from x= g(f(x)) => 1 = g'(f(x))f'(x)

    are you differentiating both sides wrt x? I'm not that familiar with the g/f version of the chain rule though I have seen it. Only in american books though. The UK doesn't seem to know about it!

    I think maybe I'm getting confused because I'm so used to seeing y is a function f of x, and not x as a function of anything.
     
  12. Jun 14, 2010 #11

    vela

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    Yes, both sides are differentiated with respect to x. It might be a little clearer to start with x=g(y), then when you differentiate, you get

    [tex]1 = \frac{dg}{dy} \frac{dy}{dx}[/tex]

    which is the form of the chain rule you're familiar with. Then since g is just x, you get

    [tex]1 = \frac{dx}{dy} \frac{dy}{dx}[/tex]
     
  13. Jun 14, 2010 #12

    HallsofIvy

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    Well, that's because the Brit's don't like to admit Leibniz had anything to do with Calculus and only use Newton's notation! :biggrin::devil:

    Newton's notation: f'(x).
    Leibniz' notation: df/dx.

    While the derivative is NOT a fraction, it is the limit of a fraction so it has "fraction properties"- Leibniz' notation allows you to think of it as a fraction. In particular, the chain rule becomes
    [tex]\frac{df}{dx}= \frac{df}{dy}\frac{dy}{dx}[/tex]
    where y is any function of x- it looks like we are just cancelling the "dy"s.

    Also if y= f(x) and, inverting, x= g(y), then
    [tex]\frac{dx}{dy}= \frac{dg}{dy}= \frac{1}{\frac{df}{dx}}= \frac{1}{\frac{dy}{dx}}[/tex]
    just "inverting the fraction" (with the understanding, again, that the derivative isn't really a "fraction", it can just be treated like one!).

     
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