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Proving an equation of f and f^(-1) derivatives

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data

    In one of my class's tests I've come across the following equation:

    [tex]\frac{d^2 y}{dx^2} \: + \: \left(\frac{dy}{dx}\right)^3 \frac{d^2 x}{dy^2} \: =0[/tex]



    2. Relevant equations

    Considering that [tex]\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}[/tex] how does one prove this statement?

    3. The attempt at a solution

    I've tried substituting like this:

    [tex]\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}= \frac{d}{dx}\left(\frac{1}{\frac{dx}{dy}}\right)= \frac{ d \left( \frac{1}{u} \right)}{du}\frac{du}{dx}=-\left(\frac{1}{\frac{dx}{dy}}\right)^2 \frac{d\left(\frac{1}{\frac{dx}{dy}}\right)}{dx}[/tex]

    but i don't see how this can continue to finalize the proof...
     
  2. jcsd
  3. Jan 25, 2012 #2

    tiny-tim

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    hi karkas! :smile:

    use the chain rule to replace that d/dx at the end by a d/dy :wink:
     
  4. Jan 25, 2012 #3
    How would that give me the needed second derivative of the inverse function??? I can't see it :(
     
  5. Jan 25, 2012 #4

    tiny-tim

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    hint: what is d/dy of 1/(dx/dy) ? :wink:
     
  6. Jan 25, 2012 #5
    Its (-d^2x/dy^2)/(dx/dy)^2 isn't it?
     
  7. Jan 25, 2012 #6

    tiny-tim

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    then doesn't that solve the question? :smile:
     
  8. Jan 25, 2012 #7
    oh man I should really learn to improve my writing...
     
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