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I'm currently studying the basics of the substitution where the author goes:

To use substitution, we write $u$ as a convenient function of $x$ ideally one for which $du$ also appears in the function we want to antidifferentiate.

For example, to antidifferentiate $x\sqrt{1+x^2}$, $\color{red}{\text{we write }u = 1 + x^2 \text{, so that } du = 2xdx}$, which implies

$$f(x)dx = \sqrt{1 + x^2}x\ dx = \frac{1}{2}\sqrt{u}\ du $$

So my main question is, where this

$$\tag{!}du = 2xdx$$ comes from?

I do understand that if $u = 1 + x^2$ then $\frac{du}{dx} = 2x$ but I don't get how we reach step $(!)$ from here.

I tried to make a wild guess by assuming that we can multiply both sides by $dx$, but seems like it would be invalid operation (or would it be?)

Stokolos Ilya
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    solve for the equation for $du$, loosely speaking, it's saying that a small change in $u$ corresponds to $2x$ multiplied by a small change in $x$ – DMH16 Dec 17 '19 at 09:58
  • By saying "solve the equation for $du$" you mean that I actually can multiply both sides of the $\frac{du}{dx} = 2x$ by $dx$? – Stokolos Ilya Dec 17 '19 at 10:00
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    Yes, it is a fraction so you can simply multiply both sides by $dx$. – DMH16 Dec 17 '19 at 10:01
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    However, I would like to stress that only the first derivatives ($du/dx$) behave like fractions of differentials. For second and higher derivatives, things do not work this way. – Vasily Mitch Dec 17 '19 at 10:03
  • @VasilyMitch This is main reason why I'm asking the question. Because, according to what I've heard so far, $\frac{dx}{dy}$ is not a fraction, but in some cases it does behave like one. The problem is, I haven't figured out what these cases are. – Stokolos Ilya Dec 17 '19 at 10:05
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    @Nevler The rigorous treatment here is to say that we have defined $du$ so that $du = \frac{du}{dx},dx$ – Ben Grossmann Dec 17 '19 at 10:05
  • @Nelver $\frac{dy}{dx}$ is never a fraction. It sometimes behaves like one because it is defined as the limit of a fraction, but any coincidental fraction-like behaviour has to be backed up by a proper proof. Unfortunately at high school level, some reasoning about $\frac{dy}{dx}$ is hand-waved and uses the fact that it sometimes behaves as a fraction in arguments to get to things like the chain rule or whatever. This is flawed reasoning, you can never assume it has certain properties unless you prove them first. – Luke Collins Dec 17 '19 at 10:13
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    For example, $1/\frac{dy}{dx} = \frac{dx}{dy}$ not because $\frac{dy}{dx}$ "behaves like a fraction", this requires a detailed proof which uses the definition of the derivative. In fact, we call this the inverse function theorem. – Luke Collins Dec 17 '19 at 10:14

3 Answers3

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If you have a function $f$ of one real variable, the differential $df$ of $f$ at $x$ is, by definition, the linear function $$df = f'(x)\, dx.$$

In your case, since $u=1+x^2$, we have $u'(x)=2x$, so $du = 2x\,dx$.

Optional: It depends how far you are in your studies of calculus that you understand why this is so by definition, and there are multiple ways to see why. The easiest is to understand that the $u$-substitution technique is summarised by the fact that $$\int_{u(a)}^{u(b)} f(x)\,dx=\int_a^b f(u(x))\,\underbrace{u'(x)\,dx}_{du},$$ which motivates that we define $du=u'(x)\,dx$.

Luke Collins
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As far as single-variable calculus goes: whenever differentials cancel out in the way that they do in a fraction, there is really a chain rule going on in the background.

For your example, we are trying to anti-differentiate $f(x) = x\sqrt{1 + x^2}$. We suspect that the antiderivative $F(x)$ is such that computing $F'(x)$ involves the chain-rule, and that the inside-function is $u(x) = 1 + x^2$. That is, we suspect that there exists some $G(x)$ such that $F(x) = G(u(x))$.

Applying the chain rule, this would imply that $$ F'(x) = G'(u(x)) \cdot u'(x) = G'(1 + x^2)\cdot 2x $$ Since we can write $f(x) = \sqrt{1 + x^2} = \frac 12 \sqrt{u(x)} \cdot u'(x)$, we can conclude that we must have $G'(u) = \frac 12 \sqrt{u}$. In other words, solving this anti-derivative now amounts to the new anti-derivative in terms of $u$, which is given by $$ G(u) = \int \frac 12 \sqrt{u}\, du. $$ This whole process of rewriting an integral is encapsulated nicely with the steps of $u$-substitution: $$ \int x \sqrt{1 + x^2}\,dx = \int \frac 12 \sqrt{u(x)} [u'(x)dx] = \int \frac 12 \sqrt{u}\,du. $$ So, it is convenient to define $du = u'(x)dx$.

Ben Grossmann
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When we think about the first derivative of one function, we can get the following conclusion:

It is supposed that $y=f(x)$, thus $$ f'\left( x \right) = \frac{{dy}}{{dx}} $$ and $$ dy = f'\left( x \right)dx, $$ which means we can treat $dy/dx$ like fractions, here is the proof $$ y' = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = f'\left( x \right) $$ so $$ \frac{{\Delta y}}{{\Delta x}} = f'\left( x \right) + o\left( {\Delta x} \right) $$ then $$ \Delta y = f'\left( x \right)\Delta x + o\left( {{{\left( {\Delta x} \right)}^2}} \right) $$ Ignore the higher-order terms $o\left( {{{\left( {\Delta x} \right)}^2}} \right)$ $$ dy = f'\left( x \right)\Delta x $$ let $dx = \Delta x$, we obtain $$ dy = f'\left( x \right)dx $$ It is noted that, the first derivative has the invariance of differential form, which means whether $u$ is an independent variable or an intermediate variable, we have $$ dy = f'\left( u \right)du. $$

Zeta
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