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Posted

[math]x=\frac{m}{2p_{x}}[/math] and [math]x^{h}=\sqrt{u\frac{p_{y}}{p_{x}}}[/math].

 

How do I show that [math]\frac{\partial{x}}{\partial{p_{x}}}=\frac{\partial{x^{h}}}{\partial{p_{x}}}-\frac{\partial{x}}{\partial{m}}x[/math]?

Posted

I forgot to add that [math]U(x,y)=xy[/math]. It's very easy now. All that needs to be done is to show that RHS=LHS.

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