calculus - $lim_{xrightarrowinfty}(frac{x+1}{x-1})^{sqrt{x^2-1}}$

I'm trying to determine $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ using L'Hopital's Rule.

I can clearly see that $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}} = \frac{\infty}{\infty},$ so we can use L'Hoptial's Rule. I'm having trouble differentiating $f(x) = (\frac{x+1}{x-1})^{\sqrt{x^2-1}}$. I've used Mathematica, but I won't understand it unless I see the step-by-step process.