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I understand the mechanics of partial and total derivatives, but the fundamental principle of the partial derivative has been troubling me for some time.

Let $x$ be a variable, and define $y := 2x$.

Define $f = x + y$. Clearly, we have that $\frac{\partial f}{\partial x} = 1$ and $\frac{\partial f}{\partial y} = 1$.

However, rewriting, we note that $f = 3x$, and so $\frac{\partial f}{\partial x} = 3$ and $\frac{\partial f}{\partial y} = 0$.

In both cases, $\frac{\text{d} f}{\text{d} x} = 3$, and so the value of $\frac{\text{d} f}{\text{d} x}$ can be unambiguously determined.

However, the value of $\frac{\partial f}{\partial x}$ appears to depend on the way that $f$ is written. Is this correct? Is it possible to unambiguously determine a single correct value of $\frac{\partial f}{\partial x}$, or can partial derivatives of the same function with respect to the same variable differ depending on how the function is written?

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  • $\begingroup$ By tying x and y together, you are effectively making f a function of one variable, x. It no longer depends on y so the partial with respect to y should be zero. The partial with respect to x is 3. The first set of partial derivatives only make sense if x and y are allowed to vary independently. $\endgroup$ – Paul Aljabar May 25 '17 at 19:00
  • $\begingroup$ $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}x+\frac{\partial}{\partial x} y = 1+\frac{\partial y}{\partial x} =1+2=3$$ $\endgroup$ – mrnovice May 25 '17 at 19:01
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The partial derivative of $f$ with respect to its first parameter is by definition found by varying that parameter while holding all others constant. By tying the two parameters together as you have, you can no longer hold the second one constant, so it doesn’t really make sense to speak of the partial derivative of this function.

What’s going on here is that you’re talking about two different functions but using the same name $f$ for both of them. The original $f$ is a function of two independent variables that happen to be called $x$ and $y$. For this function, it makes sense to speak of its partial derivatives $f_x$ and $f_y$ (or better, $f_1$ and $f_2$ so that we’re not tied to specific variable names). The second function that you’re calling $f$ is really the composition of $f$ with another function $g:x\mapsto(x,2x)$ and is really a function of only one parameter, so there aren’t any others to hold constant.

Another way to look at this is in terms of directional derivatives of $f$. The partial derivatives ${\partial f\over\partial x}$ and $\partial f\over\partial y$ are simply the directional derivatives of $f$ in the directions of the positive $x$- and $y$- axes, respectively. When you set $y=2x$, however, you restrict the domain to the line $y=2x$, and so are no longer free to move in any direction when taking derivatives of this function: you’re restricted to moving in the direction of this line, so you can’t really compute those other directional derivatives any more.

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In such a multivariable function $f(x, y)$, there is no assumption that x and y are related in any way. If you can express y in terms of x as you are trying to assume, the function becomes a 1-variable function (in terms of x or y); there is no "partial derivatives" of it. If you define $f(x, y)=x+y$, you can't change that.

Basically, by defining y=2x you're restricting the domain of the function, which you can't do.

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You are treating $y$ as a variable when it is actually a function: $y \not = y(x)$. That is $\frac{\partial y}{\partial x} = 0$ and $\frac{\partial y(x)}{\partial x}$ is some other value.

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