I'm trying to prove:

$\displaystyle \tanh(2x)=\frac{2\tanh(x)}{1+\tanh^2(x)}$

Here's what I've done so far:

$\displaystyle \tanh(2x)=\frac{\sinh(2x)}{\cosh(2x)}=\frac{\left( \frac{e^{2x}-e^{-2x}}{2}\right)}{\left(\frac{e^{2x}+e^{-2x}}{2}\right)}$

$\displaystyle =\frac{e^{2x}-e^{-2x}}{2}\frac{2}{e^{2x}+e^{-2x}}=\frac{(e^{2x}-e^{-2x})(e^{2x}+e^{-2x})+4}{2(e^{2x}+e^{-2x})}$

$\displaystyle =\frac{e^{4x}-e^{-4x}+4}{2(e^{2x}+e^{-2x})}$

From this point I'm not sure what to do. I know of the identities:

$\displaystyle \cosh(x)-\sinh(x)=e^{-x}$

$\displaystyle \cosh^2(x)+\sinh^2(x)=\cosh(2x)$

However, even with these in mind, I've been trying this for ages to no avail. Can someone give me some pointers?

Thanks