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Math Help - Implicit Differentiation problem

  1. #1
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    Implicit Differentiation problem

    Hi Everybody,

    I have a tricky implicit differentiation problem I can't seem to get my head around.

    Define the function L as follows, let's call this equation 1:

    L(a) = ( ( c - a ) / a ) X

    And define a* as the value of a which solves the following expression, let's call this equation 2:

    ( 1 - L(a*) ) ( a*) = g


    Now I am trying to calculate da* / dX which is done by differentiating both sides of equation 2. I cannot get a solution to this. My source of confusion stems from the fact that L is a function of both a* and X. Can anybody give me some guidance to this problem? Any help would be enormously appreciated. Thanks!
    Last edited by salohcin; December 17th 2012 at 01:31 PM.
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  2. #2
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    Re: Implicit Differentiation problem

    Quote Originally Posted by salohcin View Post
    Hi Everybody,

    I have a tricky implicit differentiation problem I can't seem to get my head around.

    Define the function L as follows, let's call this equation 1:

    L(a) = ( ( c - a ) / a ) X

    And define a* as the value of a which solves the following expression, let's call this equation 2:

    ( 1 - L(a*) ) ( a*) = g


    Now I am trying to calculate da* / dX which is done by differentiating both sides of equation 2. I cannot get a solution to this. My source of confusion stems from the fact that L is a function of both a* and X. Can anybody give me some guidance to this problem? Any help would be enormously appreciated. Thanks!


    L(a^*) = \frac{(c-a^*)x}{a^*}

    sub into equation (2) ...

    \left[1 - \frac{(c-a^*)x}{a^*}\right] \cdot a^* = g

    a^* - (c-a^*)x = g

    a^*(1+x) - cx = g

    I'm assuming c and g are constants ...

    \frac{d}{dx}\left[a^*(1+x) - cx = g\right]

    a^* + (1+x) \frac{da^*}{dx} - c = 0

    \frac{da^*}{dx} = \frac{c-a^*}{1+x}
    Thanks from salohcin
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  3. #3
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    Re: Implicit Differentiation problem

    This makes perfect sense! Thank you so much!
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