What does $df=g(x)dx$ mean?
I understand this $\frac{d}{dx}f(x)=g(x)$, since $\frac{d}{dx}f(x)$ has precise definition (I guess you know the definition, so I skip it.)
but what is the definition of this $df=g(x)dx$. I can "somehow" understand what it means. but does it exist a precise definition for this notation?
Thank you.
$\endgroup$ 24 Answers
$\begingroup$It is an abuse of notation, much loved especially by physicists, that means: $$ \frac{df}{dx}=g(x) $$
$\endgroup$ 1 $\begingroup$Personally, I view $df = g(x) dx$ as a mnemonic: it's formally equivalent to $\frac{df}{dx} = g(x)$, but it's written in such a way as to allow you to perform substitution of variables when integrating, i.e. $$\int_{x=\dots}^{x=\dots} (\quad\dots\quad)\; \underline{g(x)\; dx} = \int_{f=\dots}^{f=\dots} (\quad\dots\quad)\;\; \underline{df},$$ and if you think of "df" and "dx" as "a small change in f" and "a small change in x" respectively, it helps you remember how the proof goes.
$\endgroup$ $\begingroup$In Olmsted's Advanced Calculus, the author begins with a differentiable function $y=f(x)$ of $x$ and introduced two symbols, $dx$ and $dy$.
If we interpret $dx$ to be an arbitrary real number, then $dy=d(f(x))$ can be defined as function of the two independent variables $x$ and $dx$ given by
$$dy(x,dx)=f'(x)\,dx\tag 1$$
The differentials $dx$ and $dy$, as related by $(1)$, have and obvious geometric interpretation. The equation of tangent line to the curve $y=f(x)$ at $x_0$ can be written $y=f'(x_0)(x-x_0)+f(x_0)$. Then, if we change $x$ to $x+dx$, the change in $y$ is $f'(x_0)dx$, which is precisely $dy(x_0,dx)$.
$\endgroup$ $\begingroup$The second is called differential if $g(x)=f'(x)$ and represent the principal part of the change of $f(x)$ with respect to changes of $x$.
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