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Integral and Derivative Problems

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data
    2 Questions here. Please help, math test tomorrow, study group failed for these 2 problems.
    [tex]\int{\frac{1-sin\theta}{cos\theta}d\theta}[/tex]

    and

    Find [tex]\frac{dy}{dx} [/tex] if [tex]y=x^ee^x[/tex]

    2. Relevant equations
    NONE


    3. The attempt at a solution
    We've attempted every "u-substitution" we can think of for problem 1.

    We've tried to differentiate problem 2 many times, and can't come up with the proper derivative.

    thanks! Any advice is greatly appreciated.
     
  2. jcsd
  3. Jan 14, 2009 #2

    jgens

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    My recomendation for the first one is to split the integral into two separate functions. Then all you have to integrate is sec(x) - tan(x) which is quite simple.

    For the second problem show what work you've done as it should just be an application of the product rule.
     
  4. Jan 14, 2009 #3
    Thanks for the advice on Number 1! We integrated successfully!

    On to number 2 however, here's what we have so far:

    Using simply the product rule:
    [tex]y'=ex^{e-1}e^x+x^ee^x[/tex]

    We've attempted to simplify down etc, but not coming up with the right answer here!
    Is it perhaps using the idea a base besides e?

    We're really struggling here, this is our last problem!
     
  5. Jan 14, 2009 #4
    Why can't you find the right answer? Factor out e^x*x^(e-1) from both parts
     
  6. Jan 14, 2009 #5
    We're both looking at your reply with the most blank stare you can ever imagine...the [tex]x^{e-1}[/tex] is not in the second term whatsoever... Am I really as stupid as you're making me feel? I'm very confused here...
     
  7. Jan 14, 2009 #6

    jgens

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    As far as I can tell, the solution should be correct - aside from a restricted domain. Some thoughts on simplifications though: Factor (e^x)(x^e) out so you are left with [(e^x)(x^e)](e/x +1). That's about as simple as you can get it.
     
  8. Jan 14, 2009 #7
    x^e = x^(e-1) * x
     
  9. Jan 14, 2009 #8
    I would factor out the smallest of both powers which in this case is e-1 for the power of x to get

    [tex] ex^{e-1}e^x+x^ee^x = e^{x}x^{e-1}\left(e + x\right) [/tex]
     
  10. Jan 14, 2009 #9
    We opted on calling our teacher as well. And she correctly guided us through the problem... and somehow it worked. I doubt she'll give us one like this on the test... but thanks for all your help, much obliged.

    Thanks!
     
  11. Jan 14, 2009 #10

    jgens

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    Either works, though I suppose your approach yields a simpler expression.
     
  12. Jan 15, 2009 #11

    HallsofIvy

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    [itex]e^{x-1}[/itex] certainly is in the second term: the second term includes [itex]e^x[/itex] and [itex]e^x= (e^{x-1})(e)[/itex]
     
  13. Jan 15, 2009 #12
    You mean x^(e-1)?
     
  14. Jan 15, 2009 #13
    The first integral can even be done by simple substitution i.e
    [tex]
    \int{\frac{1-sin\theta}{cos\theta}d\theta}[/tex]
    can be written as,
    [tex]\int \frac{(sin {\frac {\theta}{2}}-~cos{\frac {\theta}{2}})^2}{(cos^2{\frac {\theta}{2}}-sin^2{\frac {\theta}{2}})}[/tex]
     
    Last edited: Jan 15, 2009
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