How can I find x in terms of c for this expression?

[grepecs]

Homework Statement

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.

Homework Equations

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

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

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.

 

[Mark44]

Homework Statement

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.

Homework Equations

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

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

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.

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.

 

[grepecs]

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

\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.

 

[grepecs]

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

 

[grepecs]

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