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Does the following agree with proper notation? I think my understanding is incorrect.

$$ y = f(x) $$

Without an argument ($x$ is left undefined?): $$ \frac {dy}{dx} = f' = \lim_{h \to 0} {\frac {f(h+x) - f(x)}{(h+x) -x}} $$ With an argument: $$ \frac {dy}{dx}(x)=\frac {dy}{dx} \Big|_{x} = f'(x) = \lim_{h \to 0} {\frac {f(h+x) - f(x)}{(h+x) -x}} $$ or optionally, with argument using new symbol to reduce confusion:

$$ \frac {dy}{dx}(a)=\frac {dy}{dx} \Big|_{a} = f'(a) = \lim_{h \to 0} {\frac {f(h+a) - f(a)}{(h+a) -a}} $$

Finally, I think the point of differentiation can be a variable vs argument as follows: $$ x=a,\frac {dy}{dx}= f'(a) = \lim_{h \to 0} {\frac {f(h+a) - f(a)}{(h+a) -a}} $$

My original questions are posted here and here.

Nick
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  • looks ok. But I prefer, for clarity: $ \frac {dy}{dx}(\xi)=\frac {dy}{dx} \Big|{\xi} = f'(\xi) $. Or even $\frac {dy}{dx} \Big|{x=\xi}$. – 311411 Feb 27 '21 at 15:38
  • Everything's fine, except the third Leibniz notation with the $\mid$. You can think of this line as if it's the line at the end from integrals where you have to plug the boundaries in. In this case $x$ is not a boundarie or a value at some point, it's the function variable. Also you may simplify $(x+h)-x$. – vitamin d Feb 27 '21 at 15:40
  • @311411 I thought of that earlier to eliminate confusion between $dx$ and $x$. If these are two distinct objects then there should be no confusion I think. Should there be a difference equation on the RHS of $f'$ equality? – Nick Feb 27 '21 at 15:52
  • @vitamind can you clarify? I have seen $\frac {dy}{dx}(x)=\frac {dy}{dx} \Big|{x}$ defined as such from at least one author. My understanding is the $ \Big|{x}$ represents substitution of the bar $x$ for the values of $x$ in the difference quotient or similarly substitution of $x$ into $f'(x)$. For $(x + h) - x$, I left it for understanding. – Nick Feb 27 '21 at 16:01
  • I have never seen the Leibniz notation with a bar in one "serious" paper. I would not say it's wrong it's just very uncomfortable to write and can cause confusion. – vitamin d Feb 27 '21 at 16:04
  • I had a comment that $\frac {dy}{dx} \ne f'$ and that $\frac {dy}{dx} = f'(x)$. I suspect $x=a, \frac {dy}{dx} = f'(a)$ but am not exactly sure. Does anyone know when $f'$ should be distinguished as having the argument applied as $f'(x)$? – Nick Feb 27 '21 at 21:22

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