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Derivative of x / (2 - tan x)

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to find the derivative of x / (2 - tan x).


    2. Relevant equations
    I do everything Wolfram Alpha does here:
    http://www.wolframalpha.com/input/?i=derivative+x+/+(2+-+tan+x)
    Except I don't understand why the answer is (2+x sec^2(x)-tan(x))/(-2+tan(x))^2.

    I get (x sec^2 x) / (2 - tan x).


    3. The attempt at a solution
    The solution above. Again just what Wolfram Alpha does, but I simplify:

    (2+x sec^2(x)-tan(x))/(tan(x) - 2)^2

    to

    (x sec^2 x) / (2 - tan x).
     
  2. jcsd
  3. Jul 9, 2012 #2
    You can't pull a 2-tan(x) out of the top of that equation, there is a 2-tan(x) on the top aswell as the bottom but on the top it's being added to the xsec^2(x) not multiplied so you can't simplify.
     
  4. Jul 9, 2012 #3
    Why is the tan x positive and the 2 negative in the denominator?

    [Edit] Also can't I pull 2-tan(x) out of 1(2-tan(x)) and then just leave a 1 + x sec^2(x) up there in the numerator? [Edit]
     
  5. Jul 9, 2012 #4
    I don't know why wolfram alpha changes the denominator, it has to be convention within wolfram alpha because there is no need for it. As for the numerator, it has to do with keeping track of parentheses with regards to negative signs, if you work it out it will make sense.

    As for the second comment. No, because that would require dividing 2-tan(x) out of xsec^2(x) as well, which would only make it more confusing.
     
  6. Jul 9, 2012 #5

    SammyS

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    Algebra. Algebra. Algebra !

    [itex]\displaystyle \frac{2+x \sec^2(x)-\tan(x)}{(\tan(x) - 2)^2}[/itex]
    [itex]\displaystyle =\frac{2+x \sec^2(x)-\tan(x)}{(2-\tan(x))^2}[/itex]

    [itex]\displaystyle =\frac{2-\tan(x)+x \sec^2(x)}{(2-\tan(x))^2}[/itex]

    [itex]\displaystyle =\frac{2-\tan(x)}{(2-\tan(x))^2}+
    \frac{+x \sec^2(x)}{(2-\tan(x))^2}[/itex]

    [itex]\displaystyle =\frac{1}{2-\tan(x)}+\frac{x \sec^2(x)}{(2-\tan(x))^2}[/itex]​
    This is not simpler than what we started with.
     
  7. Jul 9, 2012 #6
    Okay then thanks. I appreciate it.
     
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