Find the $n$th derivative of $f(x)=(\sqrt{x^2-1}+\sqrt{x-1})^2$ In this case $n=16$ and the point is $x=1$.
I know $\sqrt{x}$ is not differentiable in zero, but this function actually has left derivatives.
Also, I know that $f(x)=(\sqrt{x^2-1}+\sqrt{x-1})^2=x^2+x-2+2\sqrt{(x^2-1)(x-1)}$, so it is enogth to compute the derivative of $2\sqrt{(x^2-1)(x-1)}$.
Also, I tried to find the $n$th derivative of $\sqrt{g(x)}$ but its expresion was too complicated.

HINT:
Since $x\ge 1$, we may write $\sqrt{(x^2-1)(x-1)}=(x-1)\sqrt{x+1}$.  
Then, use the General Leibniz Rule for product differentiation
$$\frac{d^n}{dx^n}\left((x-1)\sqrt{x+1}\right)=\sum_{k=0}^n\binom{n}{k}\frac{d^k}{dx^k}(x-1)\frac{d^{n-k}}{dx^{n-k}}(x+1)^{1/2}$$
Finish by recognizing that the only surviving terms occur at $k=0$ and $k=1$.
