Can you solve this differentiability problem involving implicit functions?

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SUMMARY

The forum discussion centers on a differentiability problem involving implicit functions defined by the equation \(F(x,y,z)=0\). Participants were tasked with demonstrating that if \(F\) is differentiable and its partial derivatives \(F_x\), \(F_y\), and \(F_z\) are nonzero, then the relationship \(\frac{\partial z}{\partial x}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}=-1\) holds true. The problem was successfully solved by users M R, MarkFL, and Opalg, with Opalg providing a detailed solution.

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  • Knowledge of the chain rule in multivariable calculus
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Students and professionals in mathematics, particularly those studying calculus and differential equations, as well as educators looking for problem-solving techniques involving implicit functions.

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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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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: $$F_x + \frac{\partial z}{\partial x}F_z = 0$$. Hence $$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$$.

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

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