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$ xz^2+e^{yz}=0 $

I know this equation: $ 0=Fx \frac{\partial x}{\partial x} + Fy \frac{\partial y}{\partial x} + Fz \frac{\partial z}{\partial x}$

now I transform it to: $\frac{\partial z}{\partial x}=\frac{-Fx-Fy\frac{\partial y}{\partial x}}{Fz} $

and I get equation $\frac{\partial z}{\partial x}=\frac{-Fx}{Fz}=\frac{z^2}{2xz+e^{yz}y} $

Can someone explain why $ Fy\frac{\partial y}{\partial x}=0$ ?

I think when we are taking $\frac{\partial z}{\partial x} $ we are treating y as a constant so Fy=0?

user254433
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sswwqqaa
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    Pro tip: instead of $\delta$ you can use $\partial$ with the command 'partial' –  Jun 16 '17 at 17:46

1 Answers1

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Partial derivative notation is ambiguous. But you've correctly identified the crucial missing piece of information that was intended in your calculation — all of the derivatives are meant to be computed subject to the restriction that $y$ is constant.

It's not $F_y$ that's zero, however; $F_y$ still equals $ze^{yz}$. It's $\frac{\partial y}{\partial x}$ that's zero — it's zero because the intent of the calculation is to take derivatives while holding $y$ constant.

Note that if the intent was to take derivatives while holding $z$ constant, then $\frac{\partial y}{\partial x}$ would not be zero.

One notation I've seen for adding in the missing information is

$$0=F_x \left.\frac{\partial x}{\partial x}\right|_{y \text{ const}} + F_y \left.\frac{\partial y}{\partial x}\right|_{y \text{ const}} + F_z \left.\frac{\partial z}{\partial x}\right|_{y \text{ const}} $$

or some abbreviation thereof.

For the sake of precision, I should point out what $F_y$ means too. If we write down the function $F(a,b,c) = a c^2 + e^{bc}$ (so that the given equation is the value of $F(x,y,z)$), $F_y$ is shorthand for $F_2(x, y, z)$, where $F_2$ means to take the derivative of $F_2$ in its second place.

  • so if we need to calculate $\frac{\partial y}{\partial x}$ than $\frac{\partial z}{\partial x}=0$

    and the same if we want to calculate $\frac{\partial z}{\partial y}$ than $\frac{\partial x}{\partial y}=0$ ?

     – sswwqqaa Jun 16 '17 at 18:20
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    @sswwqqaa: Again, $\partial y / \partial x$ is ambiguous; since there are two degrees of freedom, it doesn't make sense to specify a derivative of one variable with respect to another unless you specify something to hold constant (unless, by chance, the two variables happen to depend on each other alone). If you need to calculate $\partial y / \partial x$ when holding $y$ constant, you get a different answer than if you need to calculate $\partial y / \partial x$ while holding $z$ constant. –  Jun 16 '17 at 18:26
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    @sswwqqaa: That said, in a setting with three variables $x,y,z$ related by one equation, if you see $\partial y / \partial x$ in isolation, holding $z$ constant is probably the intention. –  Jun 16 '17 at 18:27
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    @sswwqqaa: There's actually a notorious paradox that arises from not fully understanding this ambiguity of notation; e.g. see this stackexchange question. –  Jun 16 '17 at 18:28