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Trig Differentiation Memorization help

  1. Dec 22, 2007 #1


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    Hey guys,

    I'm having trouble memorizing the the derivatives of the trig functions. I can memorize them for a while, but I eventually forget them all after one day. Does anyone know any sort of hooks or tricks that will make this process easier?
  2. jcsd
  3. Dec 22, 2007 #2
    If you're interested in sin(x),cos(x) and tan(x):

    Firstly- it's possible to derive d(tan(x))/dx from knowledge of d sin(x)/dx and d cos(x)/dx. Unfortunately it's a little tricky to do the derivative in practice (you have to be careful):


    d tan(x)=d sin(x)/dx *cos(x)^-1 + sin(x)*d cos(x)^-1 /dx

    It's important to commit to memory that S->C and C->-S, i.e.

    d sin(x)/dx = cos(x)
    d cos(x)/dx = -sin(x)

    Unfortunately, these really have to be committed to memory- it's important to have these two derivatives at your fingertips.

    If you know complex numbers then you can derive d sin(x)/dx and d cos(x)/dx quite quickly, but I prefer to commit them to memory.
  4. Dec 22, 2007 #3
    It's probably easier to just do it with the quotient rule:

    tan(x) = sin(x) / cos(x)
    d/dx [sin(x) / cos(x)]
    = (cos(x)*cos(x) - sin(x)*-sin(x)) / (cos^2(x))
    = (cos^2(x) + sin^2(x)) / (cos^2(x))
    = 1 / cos^2(x)
    = sec^2(x)

    The other trig derivatives are simple to derive using the quotient rule.
    Last edited: Dec 22, 2007
  5. Dec 22, 2007 #4
    Also you can derive the inverse ones quite simply as well using the chain rule. For example:

    if y = arcsin(x), then
    x = sin(y) (because sin is the inverse of arcsin...). Using implicit differentiation and the chain rule,
    1 = cos(y) dy/dx
    dy/dx = 1/cos(y). So dy/dx = 1 / cos(arcsin(x)). But we can make things simpler:

    dy/dx = 1/cos(y) = 1/sqrt(1-sin^2(y)) (because cos^2(theta) + sin^2(theta) = 1, a common trig identity)
    = 1/sqrt(1-sin^2(arcsin(x)))
    = 1/sqrt(1-x^2)
  6. Dec 22, 2007 #5

    Gib Z

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    Well I would have recommended personally not to do any memorization, just *understand* these limits [tex]\lim_{h\to 0} \frac{\sin h}{h} = 1, \lim_{h\to 0} \frac{\cos h -1}{h} = 0[/tex].

    That way, from the definition of the derivative,

    [tex]\frac{d}{dx} \sin x = \lim_{h\to 0} \frac{\sin (x+h) - \sin x}{h}[/tex]
    [tex] =\lim_{h\to 0} \frac{ \sin x \cos h + \cos x \sin h - \sin x}{h}[/tex]
    [tex]= \lim_{h\to 0}\left( \sin x \cdot \frac{\cos h -1}{h} + \cos x \cdot \frac{\sin h}{h} \right)= \cos x[/tex]

    If you don't want to do any understanding...then at least remember the derivative of sin x, then you can get the derivative of cos x by the chain rule, since [tex]\cos x = \sin (\pi/2 - x)[/tex], and the derivative of tan x by the quotient rule.
    Last edited: Dec 22, 2007
  7. Dec 23, 2007 #6
    I still think it's the case that the derivatives of sin and cos really need to be committed to memory.

    Having said that- I never committed the derivative of tan or the quotient rule to memory, since they can be derived in a few seconds.

    Also- even if you do decide to commit something to memory- it's no excuse for not being able to derive it- so the above derivations are still useful.

    GibZ: what's the icon?
  8. Dec 23, 2007 #7

    Gib Z

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    It's a Blastoise, a pokemon lol.
  9. Dec 23, 2007 #8
    I'll never remember that! Do you have a derivation of it instead?
    I'm going to feel depressed if I find out GibZ is an 8 year old.
  10. Dec 23, 2007 #9


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    Why memorize them as if they were some strange, incomprehensible facts that have no rationale behind them?

    Start to UNDERSTAND maths, and sufficient memorization of the key points will develop on its own.
  11. Dec 23, 2007 #10
    It helps to keep a certain amount of information in RAM for fast access. Be honest- do you really derive dsin(x)/dx=cos(x) each and every time you come across it? Let's not be too snooty, a little memorization is a good thing if you can do it.
  12. Dec 23, 2007 #11


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    Sure, and that memorization naturally occurs once you understand the principles behind the result.

    That is why one should focus on understanding, rather than memorizing.
  13. Dec 23, 2007 #12
    That does not always follow. I understand how to derive the derivative of dtan(x)/dx but the result has never made the transition to my long term memory. If you read my above posts you'll see that I think everyone should make the attempt to derive even the results they do commit to memory. Even so, I do have some sympathy with the idea that some things are better memorized for quick recall and handy mnemonics can be a useful tool for students.
  14. Dec 23, 2007 #13


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    Active memorization has never worked for me.
    If I can't "get it", I can't remember it, however much I try.

    For that which I get, either it slides effortlessly into my memory, or I remember enough key points to derive a result when I need it.
  15. Dec 23, 2007 #14


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    First of all, you should know that sine and cosine always go into each other when taking derivatives (derivative of sine is a cosine and vice versa). If you get confused about where the minus sign goes, you can always draw a graph and recall that the derivative of a function at a certain point gives the slope of the graph at that point. Now you see that the sine starts at the origin and initially goes up, while the cosine starts at y = 1 and initially goes horizontal (and then goes down). So the derivative of the sine at 0 will be positive and as it is either +cos(0) = +1 or -cos(0) = -1 you see that
    [tex]\frac{\mathrm d}{\mathrm dx} \sin(x) = + \cos(x)[/tex]
    The cosine initially goes horizontally, which agrees with sin(0) = 0 (it has zero slope) but after a while it starts descending, so it has negative derivative there. Since the sine is initially positive, the sine which you know is in there, must be minus:
    [tex]\frac{\mathrm d}{\mathrm dx} \cos(x) = - \sin(x)[/tex]

    The tangent is just a shorthand for a specific combination of sine and cosine (namely, the quotient) so you don't have to remember it separately (and in fact, shouldn't: it might get you confused with the other two and you should know the product / quotient rule anyway so you can always derive it, as shown in this thread already).

    PS. Keep doing math, and after a while you will need these tricks less and less, and you'll even be able to immediately write down the derivative of a tangent or an arcsine without even thinking.
    Last edited: Dec 23, 2007
  16. Dec 23, 2007 #15


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    It helps to keep the small angle approximations in mind:
    sin(x) =~ x
    cos(x) =~ 1 - x^2/2

    Then you can say things like
    d/dx sin(x) =~ 1, so it must be cos(x)
    d/dx cos(x) =~ -x, so it must be -sin(x)

    Of course this is just a device to help you remember which is which. (But there is a way to make it formal and rigorous.)
  17. Dec 25, 2007 #16

    Gib Z

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    I usually just find it nicest to define trig functions in terms of their taylor series. Derivatives and small angle approximations always come very nicely from this =]
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