# What allows us divide/multiply dx in calculus?

I've read nearly all of the threads on this topic but none seem to answer my question or lead me in the best direction.

When performing U-substituion or even in it's most basic form:

$$y = 2x$$, $$dy=2\,dx$$.

What allows us to do this and what/where does in come in a calculus textbook? None of my teachers have been able to explain it to me. It makes it difficult for me to proceed when even the basics don't seem intuitive.

I've read that the idea of differentials varies but is there a single understanding regarding them to which I can pertain? I feel I'm missing something by just doing it in my calculations and not understanding why. I've genuinely read all posts regarding this, it's been something on my mind for a long time so please don't close this.

Cheers

• It's basically from the theory called Differential forms. If in short, the part we use most, is that we can do simple algebraic manipulations over them, but rule of thumb is, at the end of the day, both LHS and RHS of the relation on differential forms should have differential forms on both sides, or 0. So, for example, you can't write $dx = x$, or $dy = 1$, although you can write $dx = 0$, meaning that $dx = dt \cdot 0$. Aug 26, 2015 at 21:07
• The short answer is the chain rule. And I think the standard calculus textbook by any of thousands of authors (differing from each other in slight variations in the shade of blue used in Figure 3.52) explains that explicitly. ${}\qquad{}$ Aug 26, 2015 at 21:10
• @MichaelHardy Excuse me, what Figure 3.52?
– Vim
Aug 27, 2015 at 3:39
• @Vim Michael is being sarcastic in a roundabout, funny way: He's saying that calculus textbooks invariably cover the chain rule in more or less the same, standard manner, and that the only opportunity for those unfortunate textbooks to differentiate themselves from the others in this department are by cosmetic changes. When he nominates the blue shade of Figure 3.52 as the only place in which textbooks differ, it's not meant to be taken seriously; It's meant to be funny, by being a ridiculously specific spot and an obviously desperate attempt at trying to explain the chain rule "differently". Aug 27, 2015 at 4:12
• @IwillnotexistIdonotexist Oh thank you ;D
– Vim
Aug 27, 2015 at 4:30

The chain rule says $\dfrac d {dx} f(g(x)) = f'(g(x)) g'(x)$. Therefore

$$\underbrace{\int f'(g(x)) g'(x)\,dx = \int \left( \frac d {dx} f(g(x)) \right)\,dx}_\text{chain rule} = f(g(x)) + C.$$

If one abbreviates $g(x)$ as $u$ and $g'(x)\,dx$ as $du$, one writes this as $$\int f'(u)\,du = f(u) + C.$$

That usage is consistent with writing $\dfrac {du} {dx} = g'(x)$, as you learned to do earlier.

If your instructor couldn't tell you that this is about the chain rule (just as integration by parts is about the product rule) then either your instructor either doesn't know the material well or didn't understand what you were asking.

There is a way to make this rigorous, but it doesn't really help a calculus student to understand what's going on. Instead, this use of differentials should be regarded as a notational shorthand. It is a shorthand for the following calculation.

We know that

$$\int_a^b f'(x) \, dx = f(b)-f(a).$$

This is one side of the fundamental theorem of calculus. If we replace $$f(x)$$ with $$f(g(x))$$, then the chain rule for derivatives tells us that we must replace $$f'(x)$$ with $$f'(g(x)) g'(x)$$. Thus we expect that

$$\int_a^b f'(g(x)) g'(x) \, dx = f(g(b)) - f(g(a)).$$

Now, notice that the right side has $$f$$ evaluated at $$g(b)$$ and $$g(a)$$. That means that, taking the FTC in reverse, we get

$$\int_a^b f'(g(x)) g'(x) \, dx = \int_{g(a)}^{g(b)} f'(u) \, du.$$

This is the change of variables formula, which you might call "the reverse chain rule". In the shorthand, we replace $$g(x)$$ with $$u$$ and $$g'(x) \, dx$$ with $$du$$. The former is a proper replacement of variables, while the latter should be understood as just being notation. You can remember it through pretending that you can cancel differentials: "$$\frac{du}{dx} \, dx = du$$".

Both of the previous two equations always hold. Where change of variables does not necessarily work as expected is when the function $$g(x)$$ is not one-to-one, which (assuming $$g$$ is continuously differentiable) means that $$g'$$ has a zero.

You can see the problem by looking at an example like $$\int_{-1}^1 x^2 \, dx$$. Consider trying the substitution $$u=g(x)=x^2,g'(x)=2x$$ in this problem (even though there is obviously no need to do so). Based on the above rules, the change of variable procedure procedure says "divide by $$g'(x)$$, write what's left as some $$f(u)$$ and integrate that". So in this problem it comes down to mean "write $$\frac{x}{2}$$ in terms of $$x^2$$".

But if $$x$$ can be anything in $$[-1,1]$$, then this cannot be done with one substitution: one value of $$x^2$$ is associated to two values of $$x/2$$. Consequently you are forced to split the integral at $$0$$ to be able to make $$x/2=\sqrt{u}/2$$ on one domain and $$x/2=-\sqrt{u}/2$$ on the other (and then the procedure works properly).

• What would be a way to make this rigorous? Could you name it, so I can read up on it? Oct 3, 2016 at 16:48
• Rigorously working with infinitesimal in a theory which behaves essentially like standard analysis is the main subject of hyperreal analysis. To my knowledge all other flavors of nonstandard analysis are incompatible with standard analysis in some way.
– Ian
Oct 3, 2016 at 17:00

A simple answer is that the derivative dy/dx is the limit of Δy/Δx as Δx goes to zero. Before you take the limit, Δy and Δx are just ordinary numbers that can be multiplied or divided.

Now, consider the expression dz/dy * dy/dx . If each of the following limits exist ...

lim Δz / Δy as y → 0 = dz/dy
lim Δy / Δx as x → 0 = dy/dx
lim Δz / Δx as x → 0 = dz/dx

... then you could do the arithmetic with Δz/Δy * Δy/Δx before taking the limit, which would leave Δz/Δx.

Then, when you take the limit, you would get dz/dx.

Of course, it's important to understand that y represents a particular function of x, y=f(x), and that z represents a particular function of y, z=g(y). For a given value of x, you would have a specific value for y, which, in turn, would yield a specific value for z. In other words, these are just ordinary numbers.

But the arithmetic relationships hold no matter which value of x you choose, and no matter which functions y and z represent, provided that the limits of Δz/Δy and Δy/Δx exist. This means that the functions z(y) and y(x) must be differentiable over the entire domain of interest, and if they are differentiable, then working with the differentials is no different from working with the deltas.

There is another (not much more rigorous) explanation.

Assume that we have to calculate the integral

$$\int_a^bf(g(x))dx$$

about whose existence (in the Riemannian sense) we are convinced. Also, assume that the inverse of $g(x)$ exists over $[a,b]$.

Suppose that the substitution: $$u=g(x)$$ would help to evaluate the integral. So, let's do it. (Accept without any explanation that the new limits will be $g(a)$ and $g(b)$.)

The real question is the transformation of $dx$.

The integral above exists by hypothesis, so we can approximate it by the following Riemann sum:

$$\int_a^bf(g(x))dx\approx \sum_{i=1}^{n+1}f(g(\tilde{x_i}))(x_{i+1}-x_i).$$

Let's divide and multiply every term in the sum by

$$g(x_{i+1})-g(x_i)$$

the result is

$$\sum_{i=1}^{n+1}f(g(\tilde{x_i}))\frac{x_{i+1}-x_i}{g(x_{i+1})-g(x_i)}(g(x_{i+1})-g(x_i)).$$

The expression

$$\frac{x_{i+1}-x_i}{g(x_{i+1})-g(x_i)}$$

reminds me to the reciprocal of the derivative of $g$ at $\tilde x_i$ being the the representative point between $x_{i+1}$ and $x_i$. The shorter the interval the better the approximation is. So, let's use this intuition and change the Riemann sum to the following sum which looks to be still a valid approximation of our Riemann integral

$$\int_a^bf(g(x))dx\approx\sum_{i=1}^{n+1}f(g(\tilde{x_i}))\frac1{g'(\tilde x_i)}(g(x_{i+1})-g(x_i)).$$

Now, what is $\frac1{g'(\tilde x_i)}$? It is the derivative of the inverse of $g$ at $g(\tilde {x_i})$:

$$\frac1{g'(\tilde x_i)}=[g^{-1}]'(g(\tilde {x_i})).$$

With this invention our Riemann approximation becomes

$$\int_a^bf(g(x))dx\approx\sum_{i=1}^{n+1}f(g(\tilde{x_i}))[g^{-1}]'(g(\tilde {x_i}))(g(x_{i+1})-g(x_i)).$$

Introduce the following notations now: $u_{i+1}=g(x_{i+1})$, $u_i=g(x_i)$, and $\tilde{u_i}=g(\tilde{x_i}).$

With this we have

$$\int_a^bf(g(x))dx\approx\sum_{i=1}^{n+1}f(\tilde{u_i})[g^{-1}]'(\tilde{u_i})(u_{i+1}-u_i).$$

We recognize that the Riemann sum above is the approximation of the following integral

$$\int_{g(a)}^{g(b)}f(u)[g^{-1}](u)du.$$

That is, we have the following approximation

$$\int_a^bf(g(x))dx\approx\int_{g(a)}^{g(b)}f(u)[g^{-1}]'(u)du.$$

It is only a technical problem to show that this is not an approximation (given the conditions formulated at the beginning) but an actual equation.

Let's compare this result with the rule of integration by substitution. Again, we have $$\int_a^bf(g(x))dx$$

and we want to introduce the substitution $u=g(x)$. By the rule we take the inverse of $g$ so that we have $x=g^{-1}(u)$. Then, as the rule says we take

$$\frac{d x}{du}=\frac{d [g^{-1}](u)}{du}=[g^{-1}]'(u)$$

from where, as if it was legal, we declare that

$$dx=[g^{-1}]'(u)du$$

and we get formally the result that we gained before.

I like to think of the $$u$$ substitution as a form of reverse to implicit differentiation.

Suppose you have a function of $$u$$ namely $$F(u)$$ whose derivative with respect to $$u$$ is $$f(u)$$.

In that case you have $$\int f(u)\,du= F(u)$$.

Suppose I wished to differentiate $$F(u)$$ with respect to a variable $$x$$ where $$u=g(x)$$. We use implicit differentiation (the chain rule) to get: $$\frac{d}{dx}{F(u)}=f(u)\frac{du}{dx}$$

Now we have a situation where if we integrate the RHS with respect to $$x$$ that $$\int f(u)\frac{du}{dx}\,dx=F(u)$$. This is the reverse to implicit differentiation.

Now let's summarize and compare results: $$\color{blue}{\int} f(u) \color{blue}{du}=F(u)\\\color{red}{\int} f(u) \color{red}{\frac{du}{dx} \, dx}=F(u)$$

The blue and red integrals are equivalent and so a useful reminder is to cancel the $$dx$$

I prefer to think the $$\frac{du}{dx}$$ is a scale factor that allows integrating with respect to $$dx$$

Hopefully this helps and I haven't overlooked some vital part.

You can understand calculus very well without ever manipulating differentials at all. In fact, in a rigorous treatment of calculus (such as in Spivak's book Calculus), differentials like $dx$ have been banished (except for lingering in the notation). To understand $u$-substitution, just use the chain rule in reverse. $\int f(g(x)) g'(x) \, dx = F(g(x)) + C$, where $F$ is an antiderivative of $f$. You can check this by taking the derivative of the right hand side, using the chain rule.

• The chain rule formula in the integral is wrong, and if you take the antiderivative you get $f(g(x))+C$ (the same function). Aug 27, 2015 at 5:02
• @janesmith Note that the function I'm integrating is $f (g (x)) g'(x)$ , not $f '(g (x)) g'(x)$, so I think my formula was correct as written. Aug 27, 2015 at 5:13

When I learned calculus, I decided that $\mathrm{d}y$ means "the derivative of $y$ with respect to some variable I haven't decided upon yet".

It's a bit ad-hoc, but actually turns out to be fairly similar in spirit to rigorous methods of making sense of the notation.

In the rigorous settings, $\mathrm{d}y/\mathrm{d}x$ can be seen as shorthand for "the thing that, when multiplied by $\mathrm{d}x$, gives $\mathrm{d}y$", when that makes sense.

Your textbook probably does not actually introduce the topic properly. IMO, introductory calculus has converged to a minimalist, function-oriented framework where notations like $\mathrm{d}y$ — or even $y = 2x$ — don't really fit well, but still have to be taught anyways because not doing so would leave students unequipped and unprepared for the actual practice of calculus.