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I don't know why I find this so difficul, it seems like simple application of the chain rule but got confused somewhere. Here is the problem

I have the function $$F(W,X) = \frac{H(z)}{(W+X)^{(1-\gamma)}}$$

where $z=\frac{X}{W+X}$, $P=W+X$, $\gamma$ is a constrant and $H(z)$ is some function.

I need to find the partial derivatives $F_w, F_{wx}$ and $F_{ww}$ expressed as functions of $z,P$ and $H(z)$.

Any help is appreciated

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  • $\begingroup$ It almost seems like you are trying to change from the variables $(W,X)$ to the variables $(z,P)$. Does this mean you don't want to see $W$ or $X$ in your final expressions? $\endgroup$
    – MPW
    Commented Sep 30, 2016 at 15:32
  • $\begingroup$ yes, that's exactly right $\endgroup$
    – Daniel
    Commented Sep 30, 2016 at 15:34
  • $\begingroup$ Well I guess I'd start by writing $F = H(z) p^{\gamma - 1}$. Then I think you can say $\frac{\partial F}{\partial w} = \frac{\partial F}{\partial z}\frac{\partial z}{\partial w} + \frac{\partial F}{\partial p}\frac{\partial p}{\partial w}$. Then you should be able to manipulate the pieces to write them in terms of $z,p$ instead of $w,x$. And so on for the other derivatives. $\endgroup$
    – MPW
    Commented Sep 30, 2016 at 16:06

1 Answer 1

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Firstly consider what you have, $$F(w,x) =\frac{H(\frac{x}{w+x})}{(w+x)^{(1-\gamma)}}$$ $$z=\frac{x}{w+x},p=w+x$$ and from this, $$F(z,p) = H(z) p^{\gamma -1}$$ $$x=zp,w=p(1-z)$$

Now for $F_w$ as, $$\frac{\partial F}{\partial w} = \frac{\partial F}{\partial z}\frac{\partial z}{\partial w} + \frac{\partial F}{\partial p}\frac{\partial p}{\partial w}$$ $$\frac{\partial F}{\partial w} = p^{\gamma-1}H'(z)\frac{\partial z}{\partial w} + (\gamma-1)p^{\gamma-2}H(z)\frac{\partial p}{\partial w}$$ Here $\frac{\partial z}{\partial w}=-\frac{x}{(w+x)^2}=-\frac{zp}{(p)^2}=-\frac{z}{p}$ and $\frac{\partial p}{\partial w}=1$ so your answer for $F_w$ is, $$ F_w = -p^{\gamma-1}H'(z)\frac{z}{p} + (\gamma-1)p^{\gamma-2}H(z) $$ $$ F_w =p^{\gamma-2}( (\gamma-1)H(z)-zH'(z)) $$ And i think you can continue in this way................

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  • $\begingroup$ Thanks, this makes sense. I just think the term above should be $(\gamma -1)p^\gamma$ and not $p^{\gamma-1}$ which also asks for correction further on $\endgroup$
    – Daniel
    Commented Sep 30, 2016 at 20:20
  • $\begingroup$ @Daniel Ohh thanks i corrected it............ ( the derivative of $p^{\gamma-1}$ is $(\gamma-1)p^{\gamma-2}$ ) $\endgroup$ Commented Oct 1, 2016 at 16:04

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