There are certain quick methods called sanity checks which will catch most (but not all) arithmetic errors. One common one is to replace each number with the sum of its digits, which is the "casting out nines" method mentioned in Robert Israel's answer. To check a computation, say $567\times 894=506,898$, we replace $567$ with $5+6+7=18$ and $894$ with $8+9+4=21$, and then replace each of these with the sum of their digits to get $9$ and $3$ (in general we keep doing this until we get down to $1$ digit), while on the other side we get $5+0+6+8+9+8=36$ and then $3+6=9$. We then check that $9\times 3=9$ after casting out nines on both sides, and so our answer is probably right (though not necessarily). This method is called "casting out nines" because it ensures that whatever answer you are checking differs from the actual answer by a multiple of nine (hopefully $0\times 9$).
However, this method has a serious drawback: if the answer you are checking is correct except for having the digits switched around (a relatively common error) the method will not catch the error. A remedy for this is to use "casting out elevens" where you take the alternating sums of the digits instead of the sums, such that the last digit is always added rather than subtracted. In our previous example, this becomes $5-6+7=6$, $8-9+4=3$ and $5+0-6+8-9+8=-4$. Here we have to be a little careful: we want to take the equation $6\times 3-(-4)=0$, cast out elevens (take the alternating sum) and verify that the the resulting equation holds, which in this case it does. We move everything to one side so that we don't have to work with numbers of different signs ($6\times 3=-4$ is true $\bmod 11$, which is what matters, but it is not obvious how to cast out elevens to show this). This ensures that whatever answer you are checking differs from the actual answer by a multiple of eleven (hopefully $0\times 11$), hence the name.
Edit: These methods can both be made rigrous with modular arithmetic. The first simple checks that an equation holds $\bmod 9$, and adding the digits comes from the fact that
$$\begin{eqnarray}
d_n\cdots d_1d_0 &=& \sum\limits_{i=1}^n 10^id_i\\
&\equiv& \sum\limits_{i=1}^n 1^id_i (\bmod 9)\\
&=&\sum\limits_{i=1}^n d_i
\end{eqnarray}$$
while the second checks that an equation holds $\bmod 11$, and the alternating sum of the digits comes from the fact that
$$\begin{eqnarray}
d_n\cdots d_1d_0 &=& \sum\limits_{i=1}^n 10^id_i\\
&\equiv& \sum\limits_{i=1}^n (-1)^id_i (\bmod 11)
\end{eqnarray}$$
Credit where credit is due: I believe I read about this years ago in a question to Dr. Math from an elementary school teacher who had been teaching the method and wanted to know how it worked.
Your example makes me think of graphs.
Imagine some nice, helpful fellow came along, and made a big graph of every math concept ever, where each concept is one node and related concepts are connected by edges. Now you can take a copy of this graph, and color every node green based on whether you "know" that concept (unknowns can be grey).
How to define "know"? In this case, when somebody mentions that concept while talking about something, do you immediately feel confused and get the urge to look the concept up? If no, then you know it (funnily enough, you may be deluding yourself into thinking you know something that you completely misunderstand, and it would be classed as "knowing" based on this rule - but that's fine and I'll explain why in a bit). For purposes of determining whether you "know" it, try to assume that the particular thing the person is talking about isn't some intricate argument that hinges on obscure details of the concept or bizarre interpretations - it's just mentioned matter-of-factly, as a tangential remark.
When you are studying a topic, you are basically picking one grey node and trying to color it green. But you may discover that to do this, you must color some adjacent grey nodes first. So the moment you discover a prerequisite node, you go to color it right away, and put your original topic on hold. But this node also has prerequisites, so you put it on hold, and... What you are doing is known as a depth first search. It's natural for it to feel like a rabbit hole - you are trying to go as deep as possible. The hope is that sooner or later you will run into a wall of greens, which is when your long, arduous search will have born fruit, and you will get to feel that unique rush of climbing back up the stack with your little jewel of recursion terminating return value.
Then you get back to coloring your original node and find out about the other prerequisite, so now you can do it all over again.
DFS is suited for some applications, but it is bad for others. If your goal is to color the whole graph (ie. learn all of math), any strategy will have you visit the same number of nodes, so it doesn't matter as much. But if you are not seriously attempting to learn everything right now, DFS is not the best choice.
So, the solution to your problem is straightforward - use a more appropriate search algorithm!
Immediately obvious is breadth-first search. This means, when reading an article (or page, or book chapter), don't rush off to look up every new term as soon as you see it. Circle it or make a note of it on a separate paper, but force yourself to finish your text even if its completely incomprehensible to you without knowing the new term. You will now have a list of prerequisite nodes, and can deal with them in a more organized manner.
Compared to your DFS, this already makes it much easier to avoid straying too far from your original area of interest. It also has another benefit which is not common in actual graph problems: Often in math, and in general, understanding is cooperative. If you have a concept A which has prerequisite concept B and C, you may find that B is very difficult to understand (it leads down a deep rabbit hole), but only if you don't yet know the very easy topic C, which if you do, make B very easy to "get" because you quickly figure out the salient and relevant points (or it may be turn out that knowing either B or C is sufficient to learn A). In this case, you really don't want to have a learning strategy which will not make sure you do C before B!
BFS not only allows you to exploit cooperativities, but it also allows you to manage your time better. After your first pass, let's say you ended up with a list of 30 topics you need to learn first. They won't all be equally hard. Maybe 10 will take you 5 minutes of skimming wikipedia to figure out. Maybe another 10 are so simple, that the first Google Image diagram explains everything. Then there will be 1 or 2 which will take days or even months of work. You don't want to get tripped up on the big ones while you have the small ones to take care of. After all, it may turn out that the big topic is not essential, but the small topic is. If that's the case, you would feel very silly if you tried to tackle the big topic first! But if the small one proves useless, you haven't really lost much energy or time.
Once you're doing BFS, you might as well benefit from the other, very nice and clever twists on it, such as Dijkstra or A*. When you have the list of topics, can you order them by how promising they seem? Chances are you can, and chances are, your intuition will be right. Another thing to do - since ultimately, your aim is to link up with some green nodes, why not try to prioritize topics which seem like they would be getting closer to things you do know? The beauty of A* is that these heuristics don't even have to be very correct - even "wrong" or "unrealistic" heuristics may end up making your search faster.
Best Answer
A calculus exam is designed to test whether you know calculus, not its prerequisites. Your professor probably assumes you are well-versed enough in arithmetic and algebra that you will not make substantial "simple" errors. If you know the calculus,[*] then, but you're consistently making errors in arithmetic and algebra on the exam, then I'd suggest you need to change the focus of your studying.
Try digging up an old arithmetic or algebra textbook and working through the exercises. Do 10 or 20 at a time, then stop and grade yourself on accuracy. Keep doing this until you consistently score perfectly on these mini-exams. You're essentially drilling these skills into your subconscious so that you intuitively know how these manipulations work, and if you make a mistake, you immediately catch it.
When you're studying calculus proper, you probably go over old exam problems and homework exercises and do them. When you check your solutions to these and run across arithmetic errors, don't assume they were "just" silly mistakes and move on. You've identified that your problem on exams is exactly this, so you need to figure out where you're going wrong with the algebra/arithmetic in your problem-solving process and start working to correct it.
Here's a sports analogy. Imagine you're on a basketball team and you've got a great grasp of strategy and court sense --- you know where your teammates are, where the defenders are, who's open, and who's not. However, when you have the ball, you can't seem to control it. Your dribbling is wild and when you pass or shoot, it doesn't go where you thought you put it. In practice, what do you do? You spend hours just dribbling and then you line up and take a thousand shots at the basket. When you're playing scrimmage and the ball slips out of your hands, do you ignore it? No, you think about why you lost control and how to do better the next time. In the same way, you need to drill yourself on the prerequisites for calculus and be more self-critical when you don't understand the algebra in a problem.
[*] By this I mean that when you see a problem, you know very quickly how to solve it. For instance, if you see "Find the area between $f$ and $g$," you know immediately that you need to find the intersection points of $f$ and $g$ and the integrate $|f-g|$ between those intersection points.