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Suppose $F:\mathbb R^3\rightarrow\mathbb R$ is $C^1$, and $F(x,y,z)=0$. $z = f (x, y), x = g(y, z)$, and $y = h(x,z)$.

Show that $\frac{\partial z~\partial x~\partial y}{\partial x ~\partial y~\partial z}=-1$.

I know for $f(x,y)=0$, we get $\frac{dy}{dx}=-\frac{\frac{df}{dx}}{\frac{df}{dy}}$.

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  • $\begingroup$ I tried to use this result from $f(x,y)=0$, but then I find difficulty saying $f(x,y)=0$ given $F(x,y,z)=0$. $\endgroup$ – RRRR Apr 5 '17 at 4:44
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Differentiate the equation $F(x,y,f(x,y))= 0$ with respect to $x$ to get \begin{equation*} F_x + F_z\frac{\partial f}{\partial x} = 0. \end{equation*} Differentiate the equation $F(x, h(x,z), z) = 0$ to with respect to $z$ to get \begin{equation*} F_y\frac{\partial h}{\partial z} + F_z=0. \end{equation*} Differentiate the equation $F(g(y,z), y, z)= 0$ to wth respect to $y$ to get \begin{equation*} F_x\frac{\partial g}{\partial y} + F_y = 0. \end{equation*} Using the three computations above gives \begin{equation*} \frac{\partial f}{\partial x} \frac{\partial g}{\partial y}\frac{\partial h}{\partial z} = \left(-\frac{F_x}{F_z}\right)\left(-\frac{F_y}{F_x}\right)\left(- \frac{F_z}{F_y}\right)= -1. \end{equation*}

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  • $\begingroup$ Thank you, but I did not get why you get $F_x+F_z\frac{\partial f}{\partial x}=0$. Could you please elaborate how you differentiate that? $\endgroup$ – RRRR Apr 5 '17 at 5:42
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    $\begingroup$ The chain rule says $\frac{\partial}{\partial x} F(x,y,z) = F_x\frac{\partial x}{\partial x} +F_y\frac{\partial y}{\partial x} + F_z\frac{\partial z}{\partial x}$. Now use the relations $\frac{\partial x}{\partial x} = 1$, $\frac{\partial y}{\partial x} = 0$ and, since $z = f(x,y)$, $\frac{\partial z}{\partial x} = \frac{\partial f}{\partial x}$. $\endgroup$ – BindersFull Apr 5 '17 at 5:58

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