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Finding an expression

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    With

    [tex]f(x)=-\lgroup\frac{a}{2}-\frac{x}{2b}\rgroup\ln\lgroup\frac{a}{2}-\frac{x}{2b}\rgroup-\lgroup\frac{a}{2}+\frac{x}{2b}\rgroup\ln\lgroup \frac{a}{2}+\frac{x}{2b}\rgroup[/tex]

    and

    [tex]z=\frac{df}{dx}=\frac{1}{c}[/tex]

    find an expression of x in terms of c.

    2. Relevant equations
    Well, relevant should be that the answer is supposed to be

    [tex]x=ab \tanh\lgroup\frac{b}{c}\rgroup[/tex]

    3. The attempt at a solution

    Differentiating f wrt x, I get

    [tex]\frac{df}{dx}=-\frac{1}{2b}\lgroup \ln\lgroup\frac{a}{2}-\frac{x}{2b}\rgroup+\ln \lgroup\frac{a}{2}+\frac{x}{2b}\rgroup+2\rgroup[/tex]

    But this is no good, because when I exponentiate on both sides of the equation (in order to solve for x)

    [tex]-\frac{2b}{c}-2=-2b\frac{df}{dx}-2[/tex]

    I end up with a something like

    [tex]x=2b\sqrt\lgroup -\exp(-2\frac{b}{c}-2)-(\frac{a}{2})^2\rgroup,[/tex]

    which is obviously incorrect.

    I'd be happy if someone would help me.
     
  2. jcsd
  3. Oct 24, 2011 #2

    Mark44

    Staff: Mentor

    Did you use the product rule when you differentiated? I didn't check your work very closely, but it doesn't seem that you did.
     
  4. Oct 24, 2011 #3
    I think I did. The derivative of

    [tex]\lgroup\frac{a}{2}+\frac{x}{2b}\rgroup\ln\lgroup \frac{a}{2}+\frac{x}{2b}\rgroup[/tex]

    is, according to my calculations,

    [tex]\frac{1}{2b}\ln\lgroup \frac{a}{2}+\frac{x}{2b}\rgroup+\lgroup\frac{a}{2}+\frac{x}{2b}\rgroup\frac{1}{\frac{a}{2}+\frac{x}{2b}}\frac{1}{2b}[/tex]

    The second term reduces to 1/2b, which is broken out of the expression.
     
  5. Oct 25, 2011 #4
    No one who has any suggestions? I'm pretty sure it's a pretty simple error :)
     
  6. Oct 25, 2011 #5
    Ok, mr. Wolfram Alpha solved the problem for me (quite the dude, isn't he?).
     
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