$$\int x \sqrt[5]{3x+2} dx$$

I'm reading a book that solves this by making $t=\sqrt[5]{3x+2}$. Then it does:

$$t^5 = 3x+2\\
5t^4 dt = 3dx$$

The rest is just isolating dx, then replacing it in th expression, then taking $t^5 = 3x+2$ and doing the same for x.
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What I don't get is the step above. The book says that what was done was that the derivative was taken on both sides and that this was justified by the chain rule (I didn't get what they meant).

I played around with it and I think I figured out why, can someone tell me if I am correct? I know this notation can be treated similarly to fractions (or so i read) so here's what I did:

$$t^5 = 3x+2 \Leftrightarrow t^5\frac{1}{dt}\*\frac{1}{dx} = (3x+2)\*\frac{1}{dt}\*\frac{1}{dx} \Leftrightarrow 5t^4 \*\frac{1}{dx}=3\*\frac{1}{dt} \Leftrightarrow 5t^4dt = 3dx$$

In the last step you multiply both sides by dt and dx to get rid of the fractions.

Is this correct? If yes can someone explain to me exactly why this works?