What allows us divide/multiply dx in calculus?
Solution 1:
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.
Solution 2:
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).
Solution 3:
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.