Partial derivative notation

Question

I still have a little problem with notation for partial derivatives.

Let $$ f(x,y) = x^2y $$

What do you think that this should equal to? $$ \frac{\partial f}{\partial x}(y,x) =\, ? $$ There are two options $2yx$ or $y^2$.

Do you think that following is the same? $$ \frac{\partial f(y,x)}{\partial x}= \,? $$ And now take substitution $g(x,y)=f(y,x)$. What is following? $$ \frac{\partial g}{\partial x}(x,y)=\,? $$ I would love to hear your opinions?

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Based on DanielV comment, I need answer for things like these

$$ \frac{\partial f(y,z)}{\partial z} $$ $$ \frac{\partial f(f(y,z),z)}{\partial z} $$

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I get constantly confused during physics lectures because of this :(

Answer 1

Some terminology you should know:

• Parameters are the bound variables in function definitions. So in defining $f(x, y) = x + y + z$, $x$ and $y$ are parameters, they have no meaning outside of being placemarkers defining $f$
• Arguments are that values that are used to fill in parameters. So if I assert $\forall z~~~f(z, \pi) = f(\pi, z)$, then $\pi$ and $z$ are arguments to the function $f$ here. The scope may be global like $\pi$ or local to the expression like $z$

In common practice, when it is written

$$\frac{\partial f}{\partial x} $$

it is an abuse of notation. The idea of $x$ doesn't exist outside of defining $f$, what is indicated is the derivative of $f$ with respect to it's first argument, regardless of the name. When you see $\frac{\partial f}{\partial x} $, you should ask "where was the $x$ located when $f$ was defined?"

Physicists and engineers tend to view large problems a bit differently than mathematicians (at least those who haven't studied outside of mathematics), so things can become complicated. Real world problems tend to manifest in terms of variables like gravity, distance, time, voltage, etc, rather than functions. If you try to contort these variables and turn them into functions to conform to the calc textbooks then you are in for a world of hurt. For example, gravity could be given as a function of distance and mass, or in terms of mass and temperature where temperature is a function of distance, etc. There is no universal function that represents a variable, but variables are what must be described with the differential calculus.

So engineers and physics might actually mean

$$\frac{\partial f(y, x)}{\partial x} $$

to be interpreted as a derivative by $x$ the argument, not $x$ the parameter. You might have to ask for clarification. As mickep wisely said in the comments, it is best to avoid the reuse of variable names whenever you can. But don't forget an important point of mathematics: it isn't about presenting arguments to be memorized, it is about presenting arguments to be understood. Most ambiguous notations that occur do so because if it confuses you, then you probably aren't understanding to begin with. So before asking for clarification, first ask yourself which interpretation makes more sense in context. Then if it still isn't clear, then ask for clarification.

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So the question was updated to ask about:

$$\frac{\partial f(f(y,z),z)}{\partial z}$$

presumably under the definition $f(x, y) = x^2y$. In proper mathematical conversation, $\frac{\partial}{\partial}$ is used to refer to the derivative of a function with respect to one of its parameters. If you don't use it to mean that, then the proper folk will not invite you to their putnam exam parties.

However, in practice, sometimes something like

$$Q = \frac {\partial f(f(w,z),z)}{\partial z}$$ is used in yet another abuse of notation. This one would probably mean:

$$g(w, z) = f(f(w,z),z) \quad \text{ and } \quad Q = \frac{\partial g}{\partial z}$$

In terms of variables (rather than functions), it would mean something like "the rate of change of $f(f(w,z),z)$ with respect to $z$, assuming $\frac{dw}{dz} = 0$

The important thing to watch out for with something like this is that you can't come back later and say $w = h(v,z)$, because your use of partial derivative notation assumed this dependency didn't exist. If you add a dependency like that in later, then you have to start over from scratch with:

$$g(v,z) = f(f(h(v,z),z),z) \quad \text{ and } \quad Q = \frac{\partial g}{\partial z}$$

Answer 2

By definition, $\frac{\partial f}{\partial x}$ is a function and

$$\frac{\partial f}{\partial x} (y, x)$$

means that you plug in $(y, x)$ to evaluate your function. Thus it should be $2xy$.

Think it this way: What is

$$\frac{\partial f}{\partial x} (1, 0)?$$

Obviously you do not first plug in $x=1$ and $y=0$ and then differentiate wrt $x$.......

Answer 3

Actually,

$$\frac{\partial f}{\partial x}(y,x) $$ is the derivative over $x$ evaluating at the point $(y,x).$

Answer 4

If $f$ is defined by $f(x,y)$, the operation $\frac{\partial f}{\partial x}$ means that you differentiate $f$ with respect to $x$. With your example this will become $2xy$. Now, $\frac{\partial f}{\partial x}(y,x)$ means, that you FIRST differentiate with respect to $x$ (which was $2xy$) and AFTERWARDS plug in $(x,y)=(y,x)$. Therefore we obtain $\frac{\partial f}{\partial x}(y,x)=2yx(=2xy)$.

Now, if $g(x,y)=f(y,x)$, then $g(x,y)=y^2x$. Again, $\frac{\partial g}{\partial x}(x,y)$ means that you have to differentiate $g$ with respect to $x$ and afterwards plug in $(x,y)=(x,y)$ (so this won't change anything). Thus, we obtain $\frac{\partial g}{\partial x}(x,y)=y^2$.

Answer 5

Strictly going by the rule that in $\frac{\partial f}{\partial *}(*,\cdots,*)$, you always perform the derivative operation first and then worry about evaluation, the answer must be $2xy$. This rule must be followed, otherwise, expressions like $\frac{\partial f}{\partial y}(17)$ don't make any sense. The Leibnitzian notation is an unfortunate one to begin with and its extension to partial derivatives is bordering on nonsense. The Eulerian notation really shows its virtues in these cases. For the example above, we would write $D_1 f(y,x)$, which is completely unambiguous.