Implicit Differentiation problem

• Dec 17th 2012, 12:45 PM
salohcin
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!
• Dec 17th 2012, 01:59 PM
skeeter
Re: Implicit Differentiation problem
Quote:

Originally Posted by salohcin
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!

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

sub into equation (2) ...

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

$\displaystyle a^* - (c-a^*)x = g$

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

I'm assuming $\displaystyle c$ and $\displaystyle g$ are constants ...

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

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

$\displaystyle \frac{da^*}{dx} = \frac{c-a^*}{1+x}$
• Dec 17th 2012, 02:01 PM
salohcin
Re: Implicit Differentiation problem
This makes perfect sense! Thank you so much!