The easiest demonstration that I can think of is by using the binomial theorem. For #|x|<1# we have
# (1+x)^n = 1+nx+{n(n-1)}/{2!}x^2+{n(n-1)(n-2)}/{3!}x^3 +...#
For #(1+x)^{-1/2}# the coefficient of #x^{n-1}# will be
#{(-1/2)(-3/2)(-5/2)...(-{2n-3}/2)}/{(n-1)!}#
where the numerator has #n-1# terms. This, which can be simplified to
# (-1)^{n-1} {1cdot 3cdot 5 cdot (2n-3)}/{2^{n-1} (n-1)!}#
is the coefficient of #x^n# in #f(x) = x(1+x)^{-1/2}#.
If you now differentiate #f(x)# #n# times with respect to #x#, all powers of #x# smaller than #n# will vanish, #x^n# will yield #n!#, and all higher power terms will still have some surviving powers of #x#. If we now set #x# to zero to get #f^(n)(0)#, all these higher powers will vanish, so that we end up with
#f^(n)(0) =(-1)^{n-1} {1cdot 3cdot 5 cdot (2n-3)}/{2^{n-1} (n-1)!} times n! =(-1)^{n-1} {1cdot 3cdot 5 cdot (2n-3)}/{2^{n-1} } times n#
Direct method
We can also use Liebnitz rule of successive differentiation
#d^n/dx^n (uv) = (d^n/dx^n u)v +n (d^{n-1}/dx^{n-1} u){dv}/dx+ {n(n-1)}/{2!} (d^{n-2}/dx^{n-2} u){d^2v}/dx^2+...#
Here, if we use #u = (1+x)^{-1/2}# and #v = x#, then only the first two terms survive, and the first term vanishes at #x=0#. Thus
#f^(n)(0) = n d^{n-1}/dx^{n-1} (1+x)^{-1/2}|_{x = 0}#
Now
#d/dx (1+x)^{-1/2} = (-1/2)(1+x)^(-3/2)#
#d^2/dx^2 (1+x)^{-1/2} = (-1/2)(-3/2)(1+x)^(-5/2)#
...
#d^{n-1}/dx^{n-1} (1+x)^{-1/2} = (-1/2)(-3/2)...(-{2n-3}/2)(1+x)^(-{2n-1}/2)#
This immediately yields the result.
