How to recursively infer a word/string from a context-free grammar?

Give the recursive inference of the word $abcddd$ from the Context-free Grammar:

$A\rightarrow aAd\mid B$
$B\rightarrow bBd\mid C$
$C\rightarrow cC\mid cD$
$D\rightarrow Dd\mid ϵ$

This is a follow up question to this one here. Much like that question I haven't found any good information on the subject. I understand the inductive proof of an inferred word belonging to a language, but I don't know how to put it to practical use. I have found an example in our course literature, "Introduction to Automata Theory, Languages, and Computation" by Hopcroft, Motwani and Ullman, but they only have a filled out table of the inference of several words without any detailed explanation on how they got there.

I tried to mimic that table for this excersice and think I have solved it, but I don't understand what I'm doing or if I'm even doing it right. What I need is someone to explain the thinking behind this algorithm or whatever it is they are using. Here is the table I have created for the word $abcddd$:

$$\begin{array}{ r | c | c | c | c } & \textit{String Inferred} & \textit{For language of} & \textit{Production used} & \textit{String(s) used} \\ \hline (i) & ϵ & D & D \rightarrow ϵ & - \\ (ii) & d & D & D \rightarrow Dd & (i) \\ (iii) & cd & C & C \rightarrow cD & (ii) \\ (iv) & cd & B & B \rightarrow C & (iii) \\ (v) & bcdd & B & B \rightarrow bBd & (iv) \\ (vi) & bcdd & A & A \rightarrow B & (v) \\ (vii) & abcddd & A & A \rightarrow aAd & (vi) \\ \hline \end{array}$$

To me this is just like deriving words, but backwards, if you know what I mean? So, I am doing this correctly? Why am I doing this, instead of just deriving words "as usual"?

Labeling the productions as $A\rightarrow aAd$ (1), $A\rightarrow B$ (2), $B\rightarrow bBd$ (3), $B\rightarrow C$ (4)m $C\rightarrow cC$ (5), $C\rightarrow cD$ (6), $D\rightarrow Dd$ (6), and $D\rightarrow\epsilon$ (8) we can produce this derivation $$A\stackrel{(1)}{\Longrightarrow}aAd\stackrel{(2)}{\Longrightarrow}aBd\stackrel{(3)}{\Longrightarrow}abBdd\stackrel{(4)}{\Longrightarrow}abCdd\stackrel{(6)}{\Longrightarrow}abcDdd\stackrel{(7)}{\Longrightarrow}abcDddd\stackrel{(8)}{\Longrightarrow}abcddd$$ and reading this from right to left gives exactly the table you had.