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Inverse tan derivative

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the domain and the first derivative of 2[arctan(e^x)]



    2. Relevant equations
    d/dx arctan(x)= 1/(1+x^2)



    3. The attempt at a solution
    I'm not sure about the domain....

    For the derivative:
    d/dx 2[arctan(e^x)] = 2 [1/(1+e^x)^2] (e^x)

    But my teacher had the same answer excluding the e^x part. Isn't there supposed to be the e^x there since it is the derivative of the inside function?
     
  2. jcsd
  3. Oct 17, 2007 #2

    Dick

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    Yes, your teacher is wrong. But (1+e^x)^2 isn't the denominator. What should it be? To answer the domain question you first have to figure out the domain of arctan.
     
  4. Oct 17, 2007 #3
    Oh right, the answer should be 2 [1/1+(e^x)^2]
    The domain of arctan is (infinity, infinity)? (Since the function covers the entire graph)
     
  5. Oct 17, 2007 #4

    Dick

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    Right, so put that together with the domain of e^x. Are there any values of x where the function isn't defined?
     
  6. Oct 17, 2007 #5
    Well the graph of e^x looks like it's not defined anywhere below y=0. So would that mean the domain of e^x is x>0?
     
  7. Oct 17, 2007 #6

    Dick

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    You are confusing the domain and the range of e^x. For what values of x (not y!) is e^x defined.
     
  8. Oct 17, 2007 #7
    Oh, the domain is all reals then....so the domain of arctan(e^x) is all reals!
     
  9. Oct 17, 2007 #8

    Dick

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    Yep. I agree.
     
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