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I have a function $f(x-y)$, where $x,y \in \mathbb{R}$.

Does taking a partial derivative with respect to $x$ or $y$ make sense here?

Why would I write a function like this when I can just define $g=x-y$ and have $f(g)$ and $f’(g)$?

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    $\begingroup$ yes it make sens, but not to $f$. It make sens on $g(x,y):=f(x-y)$. $\endgroup$
    – Surb
    Jul 24 at 15:29

3 Answers 3

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Answering your Initial Question:

Does Partial Derivative make sense here ?
Definition of Partial Derivative covers the case where $f$ is a function of 2 or more variables $x,y,...$ , hence $f(x-y)$ can have Partial Derivatives & It makes sense.

Answering your Deeper Question:

Why not make it $f(g)$ where $g=x-y$ ?

There are many cases where we can indeed do that, and still extract meaningful Data.
But there are many more cases where that will lead to Data loss.

Let us say we are trying to get a relationship between Salt Deposition $f$ & the Position $(x,y)$, where there is a river running along the line $x=y$. We eventually find out that $f(x,y)=(x-y)^2$, that is, the Salt Deposition increases when moving away from the river. We can then state $f(g)=g^2$, where $g=x-y$ is the Distance from the river. All good here. The Contour Plot is given with Shading.

Now, we want to know how much Salt we can take out of the triangular area. We have to integrate, but we can not use $f(g)=g^2$ because we have lost the Data about the relationship between $x$ & $y$ & $f$ , which is necessary here.

These are very Simple Examples, we can make more complicated Examples where $f(x,y)$ is necessary & where $f(g)$ is enough.

Unsymmetrical Area

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I have a function $f(x−y)$, where $x,y∈\Bbb R$.

Does taking a partial derivative with respect to $x$ or $y$ make sense here?

Whenever $f$ is a derivable over $\Bbb R$, then $f(x-y)$ is a bivariate expression in $x,y$ with partial derivatives with respect to those variables (since $x-y$ is).

Why would I write a function like this when I can just define $g=x−y$ and have $f(g)$ and $f′(g)$?

Sure, if doing so makes it easier for you to see how to apply the chain rule, then you may do so:

$$\begin{align}\dfrac{\partial f(x-y)}{\partial x}&=\dfrac{\partial f(g)}{\partial x}&&\text{where }g:=x-y\\[1ex]&=\dfrac{\partial g}{\partial x}\cdotp\!\dfrac{\mathrm d f(g)}{\mathrm d g}\\[1ex]&= (1)\cdot f'(x-y)\\[1ex]&= f'(x-y)\\[4ex]\dfrac{\partial f(x-y)}{\partial y}&=\dfrac{\partial f(g)}{\partial y}&&\text{where }g:=x-y\text{ as before}\\[1ex]&=\dfrac{\partial g}{\partial y}\cdotp\!\dfrac{\mathrm d f(g)}{\mathrm d g}\\[1ex]&=(-1)\cdot f'(x-y)\\[1ex]&=-f'(x-y)\end{align}$$

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You can imagine that partial derivative is a directional derivative. $F(x, y)$ is a function of two-dimensions, so you have a whole plain of directions. The derivative $\partial_x$ is the derivative in the direction of the $x$-axis, and $\partial_y$ is the derivative in the direction of the $y$-axis.

Let $f$ be given. If you define $F(x,y):=f(x-y)$ and let $z=x-y$, then $f(z)=F(z+s, s)$ is true for any $s$. Now, $f'(z)=(\partial_x F(x, y))|_{(x,y)=(z+s,s)}$ for any $s$. Similarly $f'(z)=-(\partial_y F(x, y))|_{(x,y)=(s,s-z)}$ for any $s$.

Conversely, $\partial_x F(x, y) = f'(x-y)$ and $\partial_y F(x, y) = -f'(x-y)$.

So taking the partial derivative makes sense, but it can be expressed with the derivative of $f$.

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  • $\begingroup$ It is of course partially repetition of some answers already in here, so I apologize for that. Important is that you have two distinct objects, the small f and the capital F, they they are related but not identical. $\endgroup$
    – Alexisz
    Jul 26 at 3:41

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