I'm trying to come up with an expression for [itex]\partial y / \partial x[/itex] where [itex]z = f(x,y)[/itex]. By observation (i.e. evaluating several sample functions), the following appears to be true:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\begin{equation*}

\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial x} = 0

\end{equation*}

[/tex]

but I can't find anything to prove it. I can sort of get there using the differential formula

[tex]

dz = \frac{\partial z}{\partial x} \cdot dx + \frac{\partial z}{\partial y} \cdot dy

[/tex]

then dividing by [itex]dx[/itex] and reasoning that [itex]dz/dx = 0[/itex] because [itex]z[/itex] is constant when evaluating [itex]\partial y / \partial x[/itex], but I suspect this isn't a valid approach.

My questions are:

(1) Is the above relationship between [itex]\partial z / \partial x[/itex], [itex]\partial z / \partial y[/itex], and [itex]\partial y / \partial x[/itex] true?

(2) How to prove it?

P.S. Here is an example:

[tex]

\begin{align}

z = x^3y &&

y = zx^{-3} &&

\frac{\partial z}{\partial x} = 3x^2y &&

\frac{\partial z}{\partial y} = x^3 &&

\frac{\partial y}{\partial x} = -3zx^{-4} = -3x^3yx^{-4} = -3y/x

\end{align}

[/tex]

[tex]

\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial x}

= 3x^2y + x^3 \cdot \left(-3y/x\right) = 3x^2y - 3x^2y = 0

[/tex]

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# [resolved] Partial Derivative Relationships

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