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Homework Help: Derivative, but reverse?

  1. Sep 16, 2009 #1
    Okay, there are two different questions, but are answered the same way (I believe).

    1. The problem statement, all variables and given/known data
    (1)Which function has a derivative equal to ex sin 2x

    (2)Which function has a derivative equal to tan x * tan 2x * tan 3x

    3. The attempt at a solution
    I thought (since the teacher said to not use integrals) to try and solve it as an inverse function. Is this a good way to go about the problem?
  2. jcsd
  3. Sep 16, 2009 #2


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    I'm not sure what your teacher meant by "don't use integrals". The functions sought are the indefinite integrals of the given functions.

    And what do you mean by "solve it as an inverse function"?
  4. Sep 16, 2009 #3
    We are not "supposed" to know anything after, and including, the chain rule, (as of now) since we are in Calc I, and integrals are covered in Calc II.

    Since integrals and derivatives are inverses of each other, I figured I could try solving it by substituting x for y and y for x, as you would to figure out an inverse function.
    Last edited: Sep 16, 2009
  5. Sep 16, 2009 #4


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    y for x substitution won't help; the question is not asking for the inverse function, but the inverse of the derivative, which, as Halls pointed out, is the indefinite integral.

    These are both very hard questions, especially the second one (translation: I can't do the 2nd one). For the first, you know that (d/dx)e^x = e^x, and (d/dx)cos(2x)=-2 sin(2x). So the first thing to try is a e^x cos(2x), where a is a constant. But you will find that doesn't work, so try a e^x cos(2x) + b e^x sin(2x), where a and b are constants. Take the derivative, set it equal to e^x sin(2x), and solve for a and b.

    Then complain to your teacher that these are too hard. Seriously.
  6. Sep 16, 2009 #5
    Lol I know they're hard that's why I came here :P But these are bonus questions, not regular homework questions, those are easy. I'll do that for the first question, but I don't understand the constants a and b. What value do they take or do you mean just some constant which you call "a"?
    As for the second, anyone else have anything to say? If not I'll ask the teacher to do it in class :)
  7. Sep 16, 2009 #6


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    We're guessing that the function whose derivative is e^x sin(2x) takes the from f(x) = a e^x cos(2x) + b e^x sin(2x), where a and b are two constants that we don't know yet. Compute f'(x), which will obviously depend on a and b. Then see if you can find numerical values for a and b such that f'(x) = e^x sin(2x). If you can, you're done!
  8. Sep 16, 2009 #7


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    I know someone whose calc teacher forbids the use of the words "integral" before they learn "antiderivatives". Just call it an antiderivative :biggrin:
  9. Sep 16, 2009 #8
    Wow. That's like saying you aren't allowed to use the word "awesome" before learning the word "antimatter".

    Anyway. I managed to get the answer for the tangent one, but I honestly do not see how it is possible without using integral identities. Honestly, your teacher might just be messing with your head.
  10. Sep 17, 2009 #9
    Can you post the answer to it Math Jeans, I'm just curious. I'm thinking about going to his office tomorrow and asking him to work it out, because I'm interested to see how to solve it with using indefinite integrals as said before by Ivy.
  11. Sep 17, 2009 #10


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    Could you post how he does the tangent one here?
  12. Sep 17, 2009 #11
    Unless it's a cop out, sure. However, the guy has a PhD in Math, so I don't doubt his skills any.
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