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Consider the notation for denoting the differentiation of a function $f(x)$.


I mean, this notation doesn't make any sense. $dx$ means a vanishingly small $x$, which can be understood, if we take $x$ to mean $\Delta x$ in a loose sense. But what does $df(x)$ denote?

I mean, ideally, shouldn't differentiation be denoted as

$$\frac{f(x_0 + d\Delta x) - f(x_0)}{d\Delta x}$$

Is this done just to simplify things, or is there another reason?

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See this answer for an excellent summary: – Emily Feb 3 '14 at 16:20
"..if we take $x$ to mean $\Delta x$ in a loose sense." I think that it would be more accurate to say "..take $dx$ to mean $\Delta x$ in a loose sense." because loosely, $dx$ is $\Delta x$ as it approaches zero. Also, the implicit assumption in $dy/dx$ is that as $\Delta x$ approaches zero, then so does $\Delta f(x)$. Otherwise it wouldn't be differentiable. – RBarryYoung Feb 3 '14 at 16:27

The $d[f(x)]$ indicates the vanishingly small change in $f(x)$ corresponding to the vanishingly small change in $x$ indicated by $dx$. Their ratio is the derivative.

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It's a shorthand: "the derivative of a function, $f$, which is a function of $x$, with respect to $x$."

Your second definition would be


if you take the limit of that expression as $\Delta x \to 0$. Otherwise, it's more of a finite difference than a differential.

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You are close. What you are talking about is the difference quotient.$${\lim_{dx \rightarrow 0} \frac {f(x+dx)-f(x)}{dx}=\lim_{dx\rightarrow0 }\frac {\text{change in }f(x)}{\text{change in }x}}$$ . So using${f(x)=x^2}$ as an example: $${\lim_{dx \rightarrow 0}\frac{(x+dx)^2-x^2}{dx}=}$$ $${\lim_{dx \rightarrow 0}\frac{(x^2+2xdx+dx^2)-x^2}{dx}=}$$ $${\lim_{dx \rightarrow 0}\frac{2xdx+dx^2}{dx}=}$$ $${\lim_{dx \rightarrow 0} \frac {(2x+dx)dx}{dx}=}$$ $${\lim_{dx \rightarrow 0}2x+dx=}$$ $${2x}$$

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