Differential Notation Magic in Integration by u-Substitution

Question

I'm really confused now. I always thought that the differential notation $\frac{df}{dx}$ was just that, a notation.

But somehow when doing integration by u-substitution I'm told that you can turn something like this $\frac{du}{dx} = 2x\;$ into this $\;du = 2x\ dx$.

But how is that even possible? I understand that the notation comes from the fact that $\frac{du}{dx}$ actually means the limit of the difference in $u$ over the difference in $x$, with $\Delta x$ approaching $0$.

$$u'(x) = \frac{du}{dx} = \frac{du(x)}{dx} = \lim_{\Delta x\to 0} \frac{u(x+\Delta x)\ -\ u(x)}{(x+\Delta x) - x} = \lim_{\Delta x\to 0} \frac{u(x+\Delta x)\ -\ u(x)}{\Delta x}$$

So if $\frac{df}{dx}$ is just a notation for the limit mentioned above, then what is the underlying argument to say that you can treat $\frac{du}{dx}$ as if it were an actual fraction?

Appreciate the help =)

Answer 1

It is really just a notation. And the trick with the substitution e.g. $du = 2xdx$ does not have any mathematical meaning, it is just a convenient way of memorizing the integration by substitution rule/law/theorem:

$$\int_a^b f(\phi(t)) \phi'(t) dt = \int_{\phi(a)}^{\phi(b)} f(x)dx $$

Going from left to right you might want to make the substitution $x=\phi(t)$. Our mnemonic tells us to $\frac{dx}{dt} = \phi'(t)$ or in other words that you have to replace $\phi'(t)dt$ with $dx$ if you replace $\phi(t)$ with $x$. If you look again at the equation above you see that this mnemonic does a nice job, so we do not have to memorize this whole equation.

I do use the mnemonic but still I always keep this equation in mind when doing so.

Answer 2

It is just a matter of bad notations. When you see $$ \color{red}{df}=\frac{df}{dx}\color{red}{dx} $$ it does not mean that the $dx$ is cancelled out, it means that the red variables are simply ambiguous notation for something else.