Find n-th derivative of $f(x)=\sqrt{1+\cosh{x}}$ $f(x)=\sqrt{1+\cosh{x}}$
$f'(x)=\frac{\sinh{x}}{2\sqrt{\cosh{x}+1}}$
$f''(x)=\frac{(\cosh{\frac{x}{2}})^4}{(\cosh{x}+1)^\frac{3}{2}}$
I wanted to use Leibnitz rule, but i can't seem to find any similarities between derivatives. I don't know how to expand $\cosh{x}$ as Taylor series. These are the only two general methods i know for finding upper derivatives. I think there must be some trick. Anyone has any ideas?
This is what i got so far. $$f(x)=\frac{\sum_{n=0}^\infty ( \frac{(\frac{x}{2})^n}{n!}+\frac{(\frac{-x}{2})^n}{n!})}{\sqrt{2}}=\frac{\sum_{k=0}^\infty \frac{2(\frac{x}{2})^{2k}}{(2k)!}}{\sqrt{2}}$$ 
Dunno what to do now
New approach.
$f(x)=\sqrt{2}\cosh{\frac{x}{2}}$  
$f'(x)=\frac{\sqrt{2}}{2}\sinh{\frac{x}{2}}$
$f''(x)=\frac{\sqrt{2}}{4}\cosh{\frac{x}{2}}=\frac{1}{4}f(x)$
For $n$ even
$f^{(n)}=\frac{1}{2^n}f(x)=\frac{\sqrt{2}}{2^n}\cosh{\frac{x}{2}}$
For $n$ odd
$f^{(n)}=\frac{\sqrt{2}}{2^n}\sinh{\frac{x}{2}}$

$\cosh x = \frac 12 (e^x+ e^{-x})\\
\cosh^2 x = \frac 14 (e^{2x}+ e^{-2x}+1) = \frac 12 (\cosh 2x + 1)\\
2cosh^2 \frac x2 = \cosh x + 1\\
\sqrt 2 \cosh \frac x2 = \sqrt {\cosh x + 1}\\
f(x) = \sqrt 2 \cosh \frac x2$
Just wanted to verify.
The derivatives:
$f(x) = \sqrt 2 \cosh \frac x2\\
f'(x) = \frac {\sqrt 2}{2} \sinh \frac x2\\
f''(x) = \frac {\sqrt 2}{4} \cosh \frac x2\\
f^{(2k)}(x) =\frac {\sqrt 2}{2^{2k}} \cosh \frac x2\\
f^{(2k+1)}(x) =\frac {\sqrt 2}{2^{2k+1}} \sinh \frac x2$ 
