nth derivative for Taylor series of $\sqrt{1+x}$ This is a homework question, and I came to a valid answer:
$\displaystyle(1+x)^{\frac{1}{2}-n}\prod_{i=1}^{n}(\frac{3}{2}-i)$
But I am loathe to believe my professor would have given us this something like this and I feel like there is a simpler way to do this, but I just can't see it.  Is there?
 A: Not sure that this is what you are looking for but if $f(u)=u^r$ it is easy to see that:
$$f^{(n)}(u)=r(r-1)\cdot\ldots\cdot(r-n+1)u^{r-n}=n!{r\choose n}u^{r-n}$$
Then you just take $u\leftarrow 1+x$ and $r\leftarrow 
1/2$ and get your result.
A: Induction looks promising,
so I will try it.
The hypothesis is
$((1+x)^{1/2})^{(n)}
=(1+x)^{\frac{1}{2}-n}\prod_{i=1}^{n}(\frac{3}{2}-i)
$.
If $n=0$,
this is
$(1+x)^{1/2}
=(1+x)^{1/2}
$,
which is true.
If
$n=1$,
this is
$((1+x)^{1/2})'
=(1+x)^{1/2-1}(\frac32-1)
$.
The left side is
$\frac12(1+x)^{-1/2}
$
and the right side is
$(1+x)^{1/2-1}(\frac32-1)
=(1+x)^{-1/2}(\frac12)
$
so they are equal.
Suppose
$((1+x)^{1/2})^{(n)}
=(1+x)^{\frac{1}{2}-n}\prod_{i=1}^{n}(\frac{3}{2}-i)
$.
The derivative of
the left side is
$((1+x)^{1/2})^{(n+1)}
$.
The derivative of the right side is
$\begin{array}\\
(\frac{1}{2}-n)(1+x)^{\frac{1}{2}-n-1}\prod_{i=1}^{n}(\frac{3}{2}-i)
&=(1+x)^{\frac{1}{2}-(n+1)}(\frac{1}{2}-n)\prod_{i=1}^{n}(\frac{3}{2}-i)\\
&=(1+x)^{\frac{1}{2}-(n+1)}(\frac{3}{2}-(n+1))\prod_{i=1}^{n}(\frac{3}{2}-i)\\
&=(1+x)^{\frac{1}{2}-(n+1)}\prod_{i=1}^{n+1}(\frac{3}{2}-i)\\
\end{array}
$
This proves the result.
