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If $z=f(x,y)$, then total derivative is $\mathrm{d}z=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y$. If $\mathrm{d} z=0$, how do you show that $z$ is a constant?

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Since $dz=0$, $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y}=0$. Hence, on the one hand $z(x,y)=f(y)$ and on the other $z(x,y)=g(x)$ by integrating. So $f(y)=g(x)$. Differentiating both sides with respect to $x$, we have $0=\frac{\partial g}{\partial x}$. Therefore $z(x,y)=g(x)=C$.

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  • $\begingroup$ Great! that is done solving dz as an exact differential equation. Arriving at f(y)=g(x) is enough to conclude z=f(y)=g(x)=C since any expression on x alone, equal to another on y alone, can only be a constant to be well-defined. $\endgroup$ – Javier Álvarez Feb 22 '11 at 1:22
  • $\begingroup$ Right. Specifically, $g(x)-g(x_0)=f(y)-f(y)=0$ for any $x,x_0, y$. So $g(x)=g(x_0)$. $\endgroup$ – AppliedSide Feb 22 '11 at 1:40
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If $df = 0$, then $\frac{\partial f}{\partial x} \ dx = -\frac{\partial f}{\partial y} \ dy$. I guess one could solve for $f(x,y)$ to get $f(x,y) = g(x-y)$ since $(f_x+f_y)g(x-y) = g'(x-y)+(-1)g'(x-y) = 0$ identically for some $g \in C^1$.

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  • $\begingroup$ Thanks. I already know this but how does this help to show that $z$ is constant? $\endgroup$ – Vafa Khalighi Feb 21 '11 at 22:14

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