<p>Is there an easy way to remember all of the derative trig theorems like</p>
<p>dy/dx[tanx]=sec^{2}x?</p>
<p>Raw memorization!
Use note cards.
Come on, there are like 6 trig derivatives to know.</p>
<p>A couple of things that may or may not help:</p>
<p>If you're taking the derivative of a non-cofunction (sin, tan, or sec), than the deriviative of the corresponding cofunction (cos, cot, or csc) switches "co"-status and also has a negative. This relationship is probably best described in the pattern below.</p>
<p>So, d/dx[sin x] = cos x and d/dx[cos x] = -sin x
d/dx [tan x] = sec^2 x and d/dx[cot x] = -csc^2 x
d/dx [sec x] = sec x tan x and d/dx [csc x] = -csc x cot x</p>
<p>Which means if you can remember the relationship between the derivative of a trig function and the derivative of its cofunction, you only need to memorize the first entry in each row.</p>
<p>Though honestly, if you think memorizing trig derivatives is hard, wait until inverse trig functions come around. :)</p>
<p>Thanks Math Professor! That really helps! Yay, I can't wait now... :D</p>
<p>Yeah, I recommend flashcards. After a while, you just begin to know them, because you use them so much in class.</p>
<p>Conceptualise!</p>
<p>Think about the definition of a derivative.</p>
<p>Though you will loathe it at first, prove those identities over and over again, using the first equation you were given: lim as h-> 0: [f(a+h)-f(a)]/h.</p>
<p>But I assume you are on the AB syllabus -- it seems that last year there were only a couple MCQ questions that required you to know beyond the general 6 derivative trig functions.</p>
<p>Again, if you have to memorise formulas, it means that you aren't sufficiently grounded in the concept. Note cards are nice, but PROVE them! Over and over again, until you remember.</p>
<p>This is not productive if you have other topics to focus on, but if your only weak spot so far are the formulas, then this isn't too much of a waste of a time as working the whole thing out can be messy but if you step back and think for a minute at what you are doing, eventually it will be automatic.</p>
<p>Especially useful since the AP exam will cover the definition of a derivative.</p>
<p>I didn't use any memory tricks, flashcards, etc.</p>
<p>I just did a lot of problems that used the trig derivatives.</p>
<p>Just practice, practice, and practice</p>
<p>I think proving them repeatedly is probably counterproductive towards learning what the trig derivatives are. Unless it's an incentive program...</p>
<p>"I won't have to prove this again if I can just... remember it!"</p>
<p>Although proving them gives you great practice with your sum angle identities with trig functions. :)</p>
<p>Why would it? It makes you think about what you are <em>really</em> doing when taking a derivative. I think too many memorise the formulas but without grasping what the formulas are doing.</p>
<p>Do problems.</p>
<p>Why would it? It makes you think about what you are <em>really</em> doing when taking a derivative. I think too many memorise the formulas but without grasping what the formulas are doing.</p>
<p>I disagree. I can derive any formula I use pretty easily, but memorizing saves a ton of time that can be spent doing more useful things.</p>
<p>On the exam, maybe -- you definitely want the formula internalised before then.</p>
<p>But it like takes a few minutes to prove each identity -- lots of messy algebra and a bit of precal perhaps.</p>
<p>Also, you're bound to forget it next year over the summer if you simply memorise.</p>
<p>But really, the repetition of proof just becomes a mechanism for memorizing the proof... :)</p>
<p>What you are really doing when taking a derivative is finding the function that models the rate of change of the given function. The function that models the rate of change is really based upon the collection of tangent lines that model these slopes. If you understand the basic concept of what a derivative is conceptually -- for derivatives in general -- then you don't need to take that same idea and beat it upside the head with each and every single particular function to demonstrate that understanding.</p>
<p>I think both too much and too little stock is put into memorization:</p>
<p>Too much stock is put into memorization rather than learning the conceptual ideas behind the topic, especially with ideas that can easily be looked up when needed. It's much more important to know when you need the derivative of sec x and be able to look up that said derivative is sec x tan x, then to not know when you need it in the first place.</p>
<p>But at the same time, too little stock is put into memorization when we go to such extremes to avoid said memorization. On the AP Test, knowing these derivatives cold will greatly increase your chances of getting a 5, if for no other reason than that you can knock out those trig derivative questions more quickly, leaving you more time to answer other questions that you might not have otherwise gotten to.</p>
<hr>
<p>As a side note, I would venture that it takes more than just a few minutes to adequately prove each identity.</p>
<p>For instance, the proof that d/dx[sin x] = cos x is fairly complicated:</p>
<p>dy/dx
= lim (h->0) [sin (x+h) - sin x]/h
= lim (h->0) [sin x cos h + cos x sin h - sin x]/h
= lim (h->0) [sin x (cos h - 1) + cos x sin h] / h
= lim (h->0) [sin x (cos h - 1)]/h + lim (h->0) [cos x sin h]/h
= sin x * lim (h->0) [cos h - 1]/h + cos x * lim (h->0) [sin h]/h</p>
<p>Where it turns out that lim (h->0) [cos h - 1]/h = 0 and lim (h->0) [sin h]/h = 1.</p>
<p>But how do you know those? Memorize them? Evaluate for numerous values of h that are progressively closer to 0 from both sides?</p>
<p>I would venture that short of either one of these techniques that you'd have difficulty proving these statements. The idea, of course, being that you can't get away without memorizing anything.</p>
<p>Honestly, while it's nice to be able to know the proofs of the derivatives of functions like that, it's not the major point of AP Calculus AB/BC. The point is that you can apply your knowledge of derivatives to be able to answer what's at the heart of calculus: the ideas of motion and change.</p>
<p>And someone who understands conceptually what a derivative is -- in the particular case of y = sin x -- can just as easily graph Y1 = nDeriv(sin x, x, x), notice that the result is the graph of cos x, and move on in the instance where they forget the result, which ultimately, I think will prove more meaningful than reproving it using the limit definition.</p>
<p>Repetition is the key to remembering the derivatives. Keep on working on problems involving these derivatives to get them in your brain. In time, you'll be writing it in your sleep. :P</p>
<p>One way of recalling the values of trig. derivatives is to remember all the ones that start with c are NEGATIVE. Another thing to remember is that cot and tan are related respectively to csc and sec. As you can see cot and csc both start with c, so that is another way to remember.</p>
<p>Really, it is up to you to figure out how you want to memorize it. ;)</p>
<p>at least there's only six?</p>
<p>Here is a trick we were taught today in math:
Make a chart of
Sec Tan Sec
Then put the corresponding cofunctions with a negative sign in front
-Csc Cot Csc
Then to find the derivative of one of the above, you cover up the one you are looking for and read what remains of the line to find the derivative.
For example, to find the derivative of Tan, cover it up.
You are left with Sec Sec or Sec^2X or
For example, to find the derivative of CSC, Cover it up.
You are left with -CotCsc or -CscCot.
I think this is very easy to remember. The chart once again.</p>
<p>Sec Tan Sec
-Csc Cot Csc</p>
<p>Easy enough and actually useful.</p>
<p>I memorise after I've resolved what's going on. After that you find it's very easy to memorise because you know exactly why each element is there. </p>
<p>For identities, I was referring to the proofs once the basic derivative relationships between sin and cos are established. The idea is actually pretty to grasp -- during my precal tests I always used my school clock, neatly divided into 12 hours (segments of pi/3!). In precal I naturally noticed that the sin increased rather quickly then slowed down from 3 o'clock (0 radians) until 12 o'clock (pi/2); and that cos decreased rather slowly than sped up from 0 until pi/2. </p>
<p>Then later I was taught d[sinx]/dx = cos x, and that d/dx cos x = -sin x, etc. and for a while I accepted this, not really understanding what that relationship entailed (although it was pretty cool that the circular functions had ... circular derivatives that cycled on each other). Then later I was reading an ancient physics text that had a table gave cos, sin and tan values for each 0.01 of a radian up to pi/2 radians. </p>
<p>It wasn't long before I remarked, "gee, the difference between each of the sin values seems to be equivalent to the cos value beside it ... a lot of this table seems redundant, even for an era without calculators." Then I glanced down and saw that the magnitude of the rate of change [change in y-value per radian] for cos was equivalent to the sin value beside it. It was only THEN did the full significance of the idea of the derivatives of cos and sin all circularly leading to each other hit me, because how convenient is it to have two sets (of four, unless just uses the absolute value) of values whose rates of change in one list are based on the values in another, and that list's rate of change being based on the values of the first? They're like circularly defining themselves, almost like a differential equation on a different level almost, because their rates of change are based on each other.</p>
<p>Of course, this property is probably why sin and cosine are intricately connected to e, because of this property of recursively defining themselves (I mean, when one is trying to view the relationship fundamentally, and I'm not even talking about Taylor series and Euler's formula and whatnot).</p>
<p>I mean, could you create a various tables of values whose rates of change were all based on the table next, and the next, but eventually feeding back into the first one in a loop, that was not sin and cos and their respective opposites? (e^x doesn't count because except for itself it doesn't really loop once x has a coefficient more than 1)?</p>
<p>I mean, this is the stuff that I wouldn't have gotten with the formula until I thought it out for myself. I mean, I had been using "look at the clock to estimate or verify trigonometric values or trends" strategy for years but I had never put 2 and 2 together when I was first told the trig derivative formulas.</p>
<p>It's dangerous to give a formula without investigating the implications of the relationships outlined by the formula. And if you learn conceptually rather than by rote memorisation of the formulas, well you find come test day you have internalised the relationships already. </p>
<p>I just don't mean knowing how to derive the formula. You can derive the formula without knowing what the implications of the relationships are. Visualisation of what is happening as you apply the formula is key.</p>
<p>Well, the statement you made earlier was a proof using the instantaneous rate of change definition.</p>
<p>Of course, there are other "proofs" that work well as well.</p>
<p>For instance, a basic knowledge of the relationship between f and f ' will also suggest the relationship between the derivative of y = sin x and y = cos x.</p>
<p>I don't mean to imply that it's unimportant to take a look at why things work the way they do.</p>
<p>But I also think there are too many people who think they should also re-derive everything on the test, which is just as poor an idea.</p>
<p>try to make a song out of it. this may sound funny, but it actually works. in bc our math teacher makes us stand up and chant this song.</p>
<p>the derivative of sin is cooooosin and the derivative of the cosine is the negative sin. </p>
<p>tan sec squared! tan sec squared! the derivative of the tangent is the secant squared!</p>
<p>sec sec tan! sec sec tan! the derivative of the secant is the secant tan!</p>
<p>(referring to cot and csc) All derivatives of cofunctions start with a negative and they all start with co- co-!</p>
<p>hope this helps!</p>