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 $\frac{dy}{dx}=-\frac{\frac{df}{dx}}{\frac{df}{dy}}$.

  • $\begingroup$ 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

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*}

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.