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Implicit Differentiation

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Assume that the following equation define the implicit function y=(x). Find the its derivative:
    x2 + 2xy - y2 = a2
    y'=?
    y''=?

    2. Relevant equations

    [tex]\frac{dy}{dx} = -\frac{F_x}{F_y}[/tex]

    3. The attempt at a solution

    so for the first derivative I express that equation as F = x2 + 2xy - y2 - a2 = 0 and using the rule from above I get:
    [tex]y^{'} = -\frac{x+y}{x-y}[/tex] which is correct.
    For the second derivative the answer should be:
    [tex]y^{''} = \frac{2a^{2}}{(x-y)^{3}}[/tex]
    But I don't understand how to get there. Where did the 2a2 come from? the 3 hints to me that I need to make a derivative of the fraction, but I can't seem to get anything useful.
     
  2. jcsd
  3. Sep 5, 2009 #2
    you can just go through and differentiate everything w.r.t. x

    [itex]x^2+2xy-y^2=a^2[/itex]

    so, remembering to use the product rule on the second term and the chain rule on the third,

    [itex]2x+2y+2x \frac{dy}{dx}-2y \frac{dy}{dx}=0[/itex]

    then collect [itex]\frac{dy}{dx}[/itex] terms as follows:

    [itex]2(x+y)=2(y-x) \frac{dy}{dx} \Rightarrow \frac{dy}{dx}=y'=\frac{x+y}{y-x}[/itex] which is what you get.

    just do the same again to get the second derivative.
     
  4. Sep 5, 2009 #3
    Ok got it right this time.
    Just a few questions - why do you add the differential to the y and not the x?
    And in the chain rule - where is it?
    Am I correct thinking that the chain rule should be (dy/dx)(dx/dx) which cancels to dy/dx?
     
  5. Sep 5, 2009 #4

    Hootenanny

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    All latentcorpse did, was differentiate the whole expression with respect to x. Explicitly,

    [tex]\frac{d}{dx}\left(x^2+2xy-y^2\right)=\frac{d}{dx}a^2[/tex]

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

    Using the product rule on the second term and the chain rule on the final term,

    [tex] 2x + 2y\frac{d}{dx}x + 2x\frac{dy}{dx} - \frac{d}{dy}y^2\cdot\frac{dy}{dx} = 0[/tex]

    [tex]2x + 2y + 2x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0[/tex]

    Do you follow?
     
  6. Sep 5, 2009 #5
    Ok - everything is understood now. Thanks!
     
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