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Problem of the Week #71 - August 5th, 2013

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Chris L T521

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Jan 26, 2012
995
Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Suppose that the equation $F(x,y,z)=0$ implicitly defines each of the three variables $x$, $y$, and $z$ as functions of the other two: $z=f(x,y)$, $y=g(x,z)$ and $x=h(y,z)$. If $F$ is differentiable and $F_x$, $F_y$, and $F_z$ are nonzero, show that
\[\frac{\partial z}{\partial x}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}=-1.\]

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's problem was correctly answered by M R, MarkFL, and Opalg. You can find Opalg's solution below.

With $z = f(x,y)$, differentiate the equation $F(x,y,f(x,y)) = 0$ partially with respect to $x$, keeping $y$ constant and using the chain rule: \(\displaystyle F_x + \frac{\partial z}{\partial x}F_z = 0\). Hence \(\displaystyle \frac{\partial z}{\partial x} = -\frac{F_x}{F_z}\).

In the same way, \(\displaystyle \frac{\partial x}{\partial y} = -\frac{F_y}{F_x}\) and \(\displaystyle \frac{\partial y}{\partial z} = -\frac{F_z}{F_y}\). Therefore \(\displaystyle \frac{\partial z}{\partial x}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z} = -\frac{F_xF_yF_z}{F_zF_xF_y} = -1\).
 
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