(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Now, I decided for no real reason to derive a formula for the hyperbolic tangent using only what I know about the derivative of the inverse hyperbolic tangent. However, what I have looks wrong, and I'd like to check it here.

2. Relevant equations

[tex]\frac{d tanh^{-1}\left(ax\right)}{dx} = \frac{a}{1-\left(ax\right)^2}[/tex]

[tex]\frac{1}{1-u^2} = \frac{1}{2} \left(\frac{1}{u+1} - \frac{1}{u-1}\right)[/tex]

3. The attempt at a solution

So, I started with the derivative, of course...

[tex]\int \frac{a}{1-\left(ax\right)^2} dx[/tex]

I did a quick u-substitution, with u = a x and du = a dx:

[tex]\int \frac{1}{1-u^2} du[/tex]

I used the second equation in part 2, which I derived seperately:

[tex]\frac{1}{2} \left( \int \frac{du}{u+1} - \int \frac{du}{u-1} \right)[/tex]

Next came the actual integration, of course...

[tex]\frac{1}{2} \left( log\left(u+1\right) - log\left(u-1\right) \right)[/tex]

I rearranged with the logarithmic rules and resubsituted ax to get...

[tex]\frac{1}{2} log\left(\frac{ax+1}{ax-1}\right)[/tex]

Now, this should be the formula for the inverse hyperbolic tangent. So, if I set this equal to y, and then solve for x, I should have the formula for the hyperbolic tangent, right?

[tex]\frac{1}{2} log\left(\frac{ax+1}{ax-1}\right) = y[/tex]

Rearranging to isolate the logarithm...

[tex]log\left(\frac{ax+1}{ax-1}\right) = 2y[/tex]

And now I raise e to both sides to remove the logarithm.

[tex]\frac{ax+1}{ax-1} = e^{2y}[/tex]

Now, I'll multiply both sides by ax-1 (why not, what else can I do?), and see what happens. Note that from here I'm pretty much doing this as I write.

[tex]a^2 x^2 - 1 = a e^{2y} x - e^{2y}[/tex]

I'll rearrange to set the right side equal to zero...

[tex]a^2 x^2 + \left(-a e^{2y}\right) x + \left(e^{2y} - 1\right) = 0[/tex]

Now I tried to use the quadratic formula to solve this, which is probably where I went wrong...

[tex] x = \frac{a e^{2y} \pm \sqrt{a^2 e^{4y} - 4 a^2 e^{2y} + 4 a^2}}{2 a^2}[/tex]

Simplifying the radical slightly by pulling out the a, I get...

[tex]x = \frac{a\left( e^{2y} \pm \sqrt{e^{4y} - 4 e^{2y} + 4}\right)}{2 a^2}[/tex]

So, the a's will cancel, and then I recognize the radical as a perfect square...

[tex]x = \frac{e^{2y} \pm \sqrt{\left(e^{2y} - 2\right)^2}}{2a}[/tex]

And now I'll cancel the square and square root and split this into two equations...

[tex]x= \frac{e^{2y} + e^{2y} - 2}{2a} OR x = \frac{e^{2y} - e^{2y} + 2}{2a}[/tex]

And then I'll simplify that.

[tex]x = \frac{e^{2y} - 1}{a} OR x = \frac{1}{a}[/tex]

Now, neither of those seem right. What did I do wrong?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Inverse Hyperbolic Tangent

**Physics Forums | Science Articles, Homework Help, Discussion**