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This seems intuitively plausible to me. But the notation sort of gets in the way when trying to prove this exactly. In particular when using the chain rule to write $\frac{\partial}{\partial y} f(x-y) = - f'(x-y)$ the $'$ looses the information that the chain rule has already been applied.

Sebastian Oberhoff
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Let us define $u = x-y$ then we can say $$ f(x-y) = f(u). $$ This is great, but we need to transform our derivatives which becomes $$ \frac{\partial}{\partial x} = \frac{\partial u}{\partial x}\frac{d}{d u} \\ \frac{\partial}{\partial y} = \frac{\partial u}{\partial y}\frac{d}{d u} $$ respectively. We now sub in the derivatives that we multiple the $u$-derivative by as $$ \frac{\partial u}{\partial x} = 1\\ \frac{\partial u}{\partial y} = -1 $$ Thus we have $$ \frac{\partial}{\partial x} f(x-y) = 1\cdot \frac{d}{d u} f(u) = \frac{df}{du} $$ and $$ \frac{\partial}{\partial y} f(x-y) = -1\cdot \frac{d}{d u} f(u) = - \frac{df}{du} $$ or $$ -\frac{\partial}{\partial y} f(x-y) = \frac{df}{du} = \frac{\partial}{\partial x} f(x-y). $$

Chinny84
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I'm not sure that I understand your difficulty, but see if this helps. It is, I hope, clear that $$\frac{\partial}{\partial x}\sin(x-y)=\cos(x-y)$$ and $$\frac{\partial}{\partial y}\sin(x-y)=-\cos(x-y)\ .$$ The case you have described is really no different from this.

Afterthought. You say that the $f'$ "loses the information that the chain rule has been applied" - well I would say almost the opposite. It shows clearly that differentiation has been applied. OK, it isn't clear whether you used the chain rule, the product rule or whatever; but, after all, this is not really important - what is important is that you have calculated the derivative.

David
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  • I'm going over this again mentally. I guess I'm realizing just now that I never paid attention to the difference between $f'(x-y)$ and $(f(x-y))'$. – Sebastian Oberhoff Nov 29 '17 at 05:21
  • Actually it's more like $f(x-y) = f(z(x,y))$ is a single variable function in $z=x-y$, which itself is a function of two parameters $x$ and $y$, so the 'prime' can be understood as the derivative with respect to $z$, and neither $x$ nor $y$. But without proper notation this fact is not immediately obvious, which is why I don't particularly like it. – Dylan Nov 29 '17 at 05:25
  • In fact the reason 'prime' can be written as all is because $f$ is a single variable function. Otherwise if $f=f(x,y)$ then $f'(x,y)$ would of course make no sense – Dylan Nov 29 '17 at 05:29
  • Let me rephrase. I never consciously thought about the difference between $f'(g(x))$ and $(f(g(x)))'$. – Sebastian Oberhoff Nov 29 '17 at 05:30
  • @SebastianOberhoff there's no difference, both $f'(x)$ and $(f(x))'$ mean the same thing. The issue is not where the prime is placed, but with respect to which variable it's acting on. See my answer. – Dylan Nov 29 '17 at 05:44
  • I wouldn't say it's incorrect. It's just ambiguous. – Dylan Nov 29 '17 at 06:02
  • I retract my statement. They don't mean the same thing. However I maintain that it's not incorrect. Ambiguous doesn't mean incorrect, David. You gave it an interpretation that makes sense, so I don't see the problem here. – Dylan Nov 29 '17 at 06:05
  • I'd even say that you're wrong because notations like $(f(g(x))'$ or $(f(x)g(x))'$ show up all the time – Dylan Nov 29 '17 at 06:09
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We can write

$$ g(x,y) = f(z(x,y)) $$

where $z = x-y$. Then $f$ is a single-variable function in $z$ with two parameters $x,y$. Then The notation $f'(x-y)$ means something like

$$ f'(z(x,y)) = \frac{df}{dz} $$

Now if we wanted to find the derivatives $f$ w.r.t $x$ and $y$ instead, then we can employ the multivariable chain rule $$ \frac{\partial g}{\partial x} = \frac{df}{dz}\frac{\partial z}{\partial x} = \frac{df}{dz}\cdot 1 = f'(z) $$

$$ \frac{\partial g}{\partial y} = \frac{df}{dz}\frac{\partial z}{\partial y} = \frac{df}{dz}\cdot (-1) = -f'(z) $$

We can revise notation to remove the $z$ dependency and get $$ \frac{\partial}{\partial x}f(x-y) = f'(x-y) $$ $$ \frac{\partial}{\partial y}f(x-y) = -f'(x-y) $$

In this form, it can be unclear what 'prime' means. It's w.r.t to neither $x$ nor $y$ but instead an implicit variable defined as $z=x-y$.

If you're used to this short-form notation, you can derive more complicated relationships such as $$ \frac{\partial}{\partial x}f(x^2+y^2) = 2x\ f'(x^2+y^2) $$ $$ \frac{\partial}{\partial y}f(e^x\sin y) = e^x\cos y \ f'(e^x \sin y) $$

and so on.

Dylan
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