# Finding an expression

1. Oct 24, 2011

### grepecs

1. The problem statement, all variables and given/known data
With

$$f(x)=-\lgroup\frac{a}{2}-\frac{x}{2b}\rgroup\ln\lgroup\frac{a}{2}-\frac{x}{2b}\rgroup-\lgroup\frac{a}{2}+\frac{x}{2b}\rgroup\ln\lgroup \frac{a}{2}+\frac{x}{2b}\rgroup$$

and

$$z=\frac{df}{dx}=\frac{1}{c}$$

find an expression of x in terms of c.

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

$$x=ab \tanh\lgroup\frac{b}{c}\rgroup$$

3. The attempt at a solution

Differentiating f wrt x, I get

$$\frac{df}{dx}=-\frac{1}{2b}\lgroup \ln\lgroup\frac{a}{2}-\frac{x}{2b}\rgroup+\ln \lgroup\frac{a}{2}+\frac{x}{2b}\rgroup+2\rgroup$$

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

$$-\frac{2b}{c}-2=-2b\frac{df}{dx}-2$$

I end up with a something like

$$x=2b\sqrt\lgroup -\exp(-2\frac{b}{c}-2)-(\frac{a}{2})^2\rgroup,$$

which is obviously incorrect.

I'd be happy if someone would help me.

2. Oct 24, 2011

### 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.

3. Oct 24, 2011

### grepecs

I think I did. The derivative of

$$\lgroup\frac{a}{2}+\frac{x}{2b}\rgroup\ln\lgroup \frac{a}{2}+\frac{x}{2b}\rgroup$$

is, according to my calculations,

$$\frac{1}{2b}\ln\lgroup \frac{a}{2}+\frac{x}{2b}\rgroup+\lgroup\frac{a}{2}+\frac{x}{2b}\rgroup\frac{1}{\frac{a}{2}+\frac{x}{2b}}\frac{1}{2b}$$

The second term reduces to 1/2b, which is broken out of the expression.

4. Oct 25, 2011

### grepecs

No one who has any suggestions? I'm pretty sure it's a pretty simple error :)

5. Oct 25, 2011

### grepecs

Ok, mr. Wolfram Alpha solved the problem for me (quite the dude, isn't he?).