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I started Calculus 2 on Monday, where we're beginning with Integration. As somewhat of a refresher, the professor is having us find the derivative of several equations and writing them in the form

$$\frac{d}{dx}(x^2-4x)=2x-4 \to d(x^2-4x)=(2x-4)dx$$

I understand that this has something to do with preparing us to understand why you see integrals in the form of

$$\int(2x-4)dx = x^2-4x+C$$

but I was led to believe that you can't do this, because $\frac{d}{dx}$ is not a fraction.

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Take a look at the FAQ. –  Fabian Jan 9 '13 at 16:29
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3 Answers

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Think of it this way:

$$ \frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$$

and, as $\Delta y$ and $\Delta x$ are real numbers, then by multiplying by $dx$, we are really just multiplying by, roughly, $\Delta x$. It's OK.

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The right question to ask is what $df$ and $dx$ mean in $$d(x^2-4x)=(2x-4)dx?$$ $\frac{df}{dx}$ is not a fraction but rather a symbol, while $df$ and $dx$ are "entities-symbols" that only make sense in $\frac{df}{dx}$ and not in expressions as above unless one gives them a standalone meaning (see Differential forms) .

Now for Calculus 2, these two symbols are (usually) not rigorously defined and so the statement $$d(x^2-4x)=(2x-4)dx$$ makes little to no sense at all. Such non-rigorous manipulations however simplify many concepts in integration:

Integration by Substitution the rigorous way:

Let $g:[a,b]\to \mathbb{R}$ be a differentiable function such as that $g^{\prime}$ is integrable. If $I=g([a,b])$ and $f:I\to \mathbb{R}$ is continuous then \begin{equation} \int\limits_{a}^{b} (f\circ g)g^{\prime}= \int\limits_{g(a)}^{g(b)} f\end{equation}

To make the above method simpler to state one uses $du$ and $dx$ separately:

Integration by Substitution the non-rigorous way:

Say we want to evaluate, \begin{equation} \int\limits_{a}^{b} f(g(x))g^{\prime}(x)\, dx\end{equation} Setting $u=g(x)$ then $du=g^{\prime}(x)dx$ and for $x=a$, $u=g(a)=k$, for $x=b$, $u=g(b)=l$. Thus, \begin{equation} \int\limits_{a}^{b} (f(g(x)))g^{\prime}(x)\, dx=\int\limits_{k}^{l} f(u)\, du\end{equation}

Simpler than before isn't it?

So to conclude, you are not really doing rigorous math by multipliying by "$dx$", rather you play with the symbols and simplify the results.

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df and dx are one-forms. This basically means they act on (tangent) vectors to give real numbers: they are 'dual' to them. So in this sense, they are real 'things' that we can manipulate on their own - the differential df of f, at a point x_0, is defined as a certain linear map of the variable dx.

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He started Calculus 2 on Monday (last January) ... are you sure he is ready for 1-forms? –  GEdgar Apr 7 '13 at 21:12
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