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Finding an expression

  • Thread starter grepecs
  • Start date
  • #1
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Homework Statement


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.

Homework Equations


Well, relevant should be that the answer is supposed to be

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

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.
 

Answers and Replies

  • #2
33,270
4,975

Homework Statement


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.

Homework Equations


Well, relevant should be that the answer is supposed to be

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

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.
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.
 
  • #3
17
0
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.
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.
 
  • #4
17
0
No one who has any suggestions? I'm pretty sure it's a pretty simple error :)
 
  • #5
17
0
Ok, mr. Wolfram Alpha solved the problem for me (quite the dude, isn't he?).
 

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