Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Take a look at the FAQ. – Fabian Jan 9 '13 at 16:29
up vote 2 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.