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Help with derivates

  1. Sep 3, 2010 #1
    1. The problem statement, all variables and given/known data

    I need help differentiating these expressions. I'm taking Calc II right now so I'm reviewing the stuff I forgot how to do.

    1.) eln2x

    2.) sin6xcsc6x

    3.) sin2x/3 + cos2x/3

    3. The attempt at a solution

    1.) This is 2. I know the ln cancels the power so it's simplified to 2x. What if the ln and e switched places? Would it still cancel out to 2x?

    2.) I have no idea how to approach this problem. The answer is supposed to be 0. Do I use the product rule?

    3.) I also don't know how to approach this problem. Are there any trig identities needed to know on how to solve this? The answer should be 0.
  2. jcsd
  3. Sep 3, 2010 #2


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    Remembering your trig identities would certainly help you do these problems quickly. But they are certainly unnecessary for simply doing the problems; you can just differentiate like you normally would.
  4. Sep 5, 2010 #3
    I'm not getting them. I'm trying to use the product rule for 2. I get an answer of 6cos6xcsc6x - 6csc6xcot6xsin6x. I must be doing something wrong because I have no idea how that would simplify to 0.

    For number 3, I get -6x-2sin(x/3) + -6x-2cos(x/3)
  5. Sep 5, 2010 #4


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    Your result for #2 looks accurate; I'll assume you did the work correctly.

    While it doesn't prove it simplifies to zero, you could always plug in values for x to see if it evaluates to zero, to gain a little bit of confidence in it.

    Incidentally, when simplifying trig functions, I often find it convenient to turn everything into sines and cosines -- or even just sines (or just cosines) if possible. This removes a lot of redundancy, and the algebraic identities satisfied by the sine and cosine are, IMO, a lot simpler and easier to recognize/apply than the other ones.

    For #3, could you show your work? I'm having great trouble trying to guess what you did.
  6. Sep 5, 2010 #5
    If you were to switch the ln and the e I don't believe they would cancel out. If you are ever unsure about this property just realize that if you have y = ex then x = lny, so if you sub this back into the original you are left with y = elny which basically proves the statement. That's the way I always like to think about it.​
    Like Hurkyl said, you should definitely look at the the trig identities, they will make your life much easier and in the long run there will be less things you have to remember because all the trig identities are really just based on the identities that deal with sin, cos, and tan.​
    Like with #2 I recommend using the trig identities. However if you want to push through it and differentiate, you must remember to use the chain rule. If, for example you had a function and it was f(x) = cos2(x3/3), then our derivative would be -2x2cos(x3/3)sin(x3/3). I hope that refreshes your memory to the chain rule. ​

    NOTE: It's extremely importannt to know at least the basic trig identities (you can work your way to the more difficult ones from these) so here is a link: http://www.sosmath.com/trig/Trig5/trig5/trig5.html. I suggest you look through them, but don't necessarily memorize them (there's a lot there). Instead remember the first 2 parts, and then prove the rest of them. Reason I say this is because (I'm assuming you're doing Calc II in uni.) in Calc II or any future calc courses simplifying is going to save you a LOT of time.
  7. Sep 5, 2010 #6
    e^ln(x) = x ;
    ln(e^x) = x*ln(e) = x ;

    csc(x) = 1/sin(x) ;
    sec(x) = 1/cos(x) ;

    cos(x)^2 + sin(x)^2 = 1 ;
    1 + tan(x)^2 = sec(x)^2 ;
    cot(x)^2 + 1 = csc(x)^2

    *be sure you memorized the above!
    good luck :)
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