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Derivatives of Exponentials

  1. Jun 6, 2005 #1
    Hi all just studying up for my exam :grumpy: im scared,

    ive been going over some practice questions and im stuck with a couple and would like some better understanding of another.

    first of all:
    determine the derivatives of each of the following functions
    (1) x^3 e^2x

    with this i know that the answer is x^2(3+2x)e^2x. i got the answer right but even im not sure how, could someone explain to me the (3+2x) in the brackets, i dont understand it because i know d/dx e^x = e^x and d/dx e^f(x) = e^f(x) x f '(x).

    (2) e^(x^2+1)^1/2

    with this one i tried using the chain rule for (x^2+1)^1/2 to get the derivative, i got what i thought was the right answer but wasnt. what am i doing wrong here? am i using the right rule?

    (3) (e^2x - 2e^x)^2

    once again i tried the chain rule here ended up getting close to the answer but it all seemed to cancel each other out? i belive i got close to this one but iam missing something. once again am i using the right rule?

    (4) e^x + e^-x / e^x - e^-x

    i tried the quotient rule for this one but i pretty much ended up with the same but with (e^x - e^-x) on top with e^x + e^-x.

    any help would be appreciated thanks guys
  2. jcsd
  3. Jun 6, 2005 #2


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    Alright,let's take it methodically.

    For the first exercise,i have a hunch that "(3+2x)" comes from that simple algebra operation called "factoring of common parts" which comes from the distributivity of the multiplication wrt addition.

    For the second,it's just chain rule.You must be screwing up that sqrt's derivative.

    For the third,hmm,i think it's easier than the second.

    The fourth is simply [itex] \left(\coth x\right)' [/itex].It's "-1" times "hyperbolic cosecant squared".

  4. Jun 6, 2005 #3
    sorry still not on the ball, would you care to explain in a little more depth?

    and thank you for the help on the other 3.
  5. Jun 6, 2005 #4


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    [tex] \left(x^{3}e^{2x}\right)'=3x^{2}e^{2x}+2x^{3}e^{2x}=x^{2}\left(3+2x\right) e^{2x} [/tex]

    I say it's not that difficult,is it...?

  6. Jun 6, 2005 #5
    ohhh i get ya :blushing: sorry
  7. Jun 6, 2005 #6


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    We'd like to think introductory algebra as a prerequisite for calculus.So my advice is to brush on some rusty simple theorems in elementary algebra.

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