Explain $du = g'(x) dx$ when $u=g(x)$ I have seen the following multiple times in the book Calculus by James Stewart.

For instance, it appears in the proof of the substitution rule (or change of variable) for integral.

Suppose $u = g(x)$ and $\dfrac{du}{dx} = g'(x)$. 

Then, $du = g'(x) dx$.

$\dfrac{du}{dx}$ is the derivative of $u$ with respect to $x$ not it is definitely not a fraction.

For instance, the $\dfrac{dy}{dx} = \dfrac{dy}{dt}\times \dfrac{dt}{dx}$ holds true as proven as the Chain Rule.

So why does $du = g'(x) dx$ hold  true? And what does this mean?

I need a clear explanation on this concept please. Appreciate your help, thanks :)
The Leibniz notation does not denote a fraction, as you know, so it is not technically correct to treat it as one and 'separate' the top and bottom. Also, given $u = g(x)$ we never need to say things like $du = g'(x) dx$. After all, simple analysis suffices to get what you actually want, in this case the change of variable theorem.
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\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$



Take any differentiable function $g : D \to U$, and any real/complex parameter $x$ varying in $D$.

Let $u = g(x)$. Then $\lfrac{du}{dx} = g'(x)$.

Take any continuous function $f$ on $U$. Then $\int f(u)\ du$ exists (on the path traced by $x$).

Thus $\lfrac{d(\int f(u)\ du)}{dx} = \lfrac{d(\int f(u)\ du)}{du}·\lfrac{du}{dx} = f(u)\lfrac{du}{dx}$, by chain rule and definition of anti-derivative.

Thus $\int f(u)\ du = \int f(u) \lfrac{du}{dx}\ dx + k$ for some constant $k$. Note that here we are using a basic theorem that if $\lfrac{dy}{dx} = z$ everywhere (on the path traced by $x$) then $y = \int z\ dx + k$ for some constant $k$.

You can see that nowhere do we break apart the derivative notation.



To look at the little bits, recall that as $Δx \to 0$ we have $\lfrac{Δu}{Δx} \approx g'(x)$, which implies much more than $Δu \approx g'(x)·Δx$. It in fact implies that $Δu - g'(x)·Δx \ll Δx$.

Intuitively, as $x$ changes by $Δx$, $u$ will change by $Δu$, and the change in $\int f(u)\ du$ divided by $Δu$ is approximately $f(u)$. Hence $Δ(\int f(u)\ du)$ divided by $Δx$ is approximately $f(u)·\lfrac{Δu}{Δx}$. If you add up the little bits over the entire path of $x$, then the total change of $\int f(u)\ du$ is approximately the sum of $f(u)·\lfrac{Δu}{Δx}$ over all the little changes in $x$.
When you are working with differentials you actually zoom so close to point on the curve that the curve behaves as a straight line.



Coefficient of the line is derivative of $g(x)$ in that point.

So yes, it is actually a fraction.