Calculating a derivative of (2015)-th order of function $f(x)$ So I have
$\displaystyle{\,\mathrm{f}\left(\, x\, \right) =
\frac{x - 1}{\,\sqrt{\, 3 - x\,}\,}}$
And I want to calculate $\,\mathrm{f}^{\mathrm{\left(\, 2015\, \right)}}
\,\left(\, 1\,\right)$
So I got the first and second derivative:
$$\mathrm{f}'(x)=\frac{-x+5}{2[(-x+3)^{\frac{3}{2}}]}$$
$$ \mathrm{f}''(x)= \frac{-2(-x+3)^{\frac{3}{2}} + (-x+5)\sqrt{-x+3}}{4(-x+3)^3}$$
Perhaps I should look at some of the next derivatives for pattern? or is that not going to help, there is also a formula I know for getting a high order derivative, but it's not going to help if I don't calculate all 2014 of them as well...
So any help with this would be appreciated.
 A: Hint $$f(x)= \frac{x-1}{\sqrt{3-x}}=\frac{2}{\sqrt{3-x}}-\frac{3-x}{\sqrt{3-x}}=2(3-x)^{-\frac{1}{2}}-(3-x)^{\frac{1}{2}}$$
A: If one knows the following standard Taylor series expansion,

$$
\frac{1}{\sqrt{1-t}}=\sum_{n=0}^\infty \frac{{2n \choose n}}{2^{2n}}\:t^n, \quad |t|<1, \tag1
$$ 

then one gets
$$
f(x)= \frac{x-1}{\sqrt{3-x}}=\frac1{\sqrt{2}}\cdot\frac{x-1}{\sqrt{1-\frac{x-1}2}}=\frac1{\sqrt{2}}\cdot\sum_{n=0}^\infty \frac{{2n \choose n}}{2^{3n}}\:(x-1)^{n+1}
$$ giving

$$
\frac{f^{(2015)}(1)}{2015!}=\frac1{\sqrt{2}}\:\frac{{4028 \choose 2014}}{2^{6042}}. \tag2
$$

A: Using Leibniz Formula for higher derivatives
$$((x-1)\cdot (3-x)^{-1/2})^{(2015)}=\sum_{k=0}^{2015}{2015 \choose k}(x-1)^{(k)}((3-x)^{-1/2})^{(2015-k)}.$$
Clearly the only summand that survives  are the ones with $k=0,1$ as for higher orders $(x-1)^{(l)}=0$. Moreover for $x=1$ the $k=0$ the term is also zero because $1-1=0$. So the only remaining term is
$$f^{(2015)}(1)=((x-1)\cdot (3-x)^{-1/2})^{(2015)}\Big|_{x=1}={2015 \choose 1}(x-1)^{(1)}((3-x)^{-1/2})^{(2015-1)}\Big|_{x=1}\\ ={2015 \choose 1}(-1)^{2014}(3-x)^{-1/2-2014}4027!!/2^{2014}\Big|_{x=1}\\ =2015(-1)^{2014}4027!!2^{-(1/2+4028)}$$.
A: Hint. Observe that
$$
\frac{x-1}{\sqrt{3-x}}=-\sqrt{3-x}+\frac{2}{\sqrt{3-x}}.
$$
Then, show inductively that, for $n\ge 2$
$$
\frac{d^n}{dx^n}\sqrt{3-x}=-\frac{(2n-3)!}{2^{2n-2}(n-2)!}(3-x)^{-\frac{2n-3}{2}}
$$
A: $$
f(x)=\frac{x-1}{\sqrt{3-x}}
$$
set $t=x-1$:
$$
f(t)=\frac{t}{\sqrt{2-t}}=\frac{1}{\sqrt{2}}t(1-t)^{-\frac{1}{2}}
$$
now use the taylor series for $(1+t)^\alpha$ (find it here):
$$
f(t)=\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}\frac{t^{n+1}}{\sqrt 2}
$$
now change index from $n+1\to n$:
$$
f(t)=\sum_{n=1}^{\infty}\binom{-\frac{1}{2}}{n-1}\frac{t^{n}}{\sqrt 2}
$$
so by definition of Taylor series the coefficient in front of the k-th power is the k-th derivative divided by k!, hence:
$$
f^{(k)}_{x=1}=\binom{-\frac{1}{2}}{k-1}\frac{k!}{\sqrt 2}=\frac{2\cdot k!}{2^k\sqrt 2}\prod_{j=1}^{k-1}\frac{1-2j}{j}=\frac{(-1)^{k+1}2k}{2^k\sqrt 2}\prod_{j=1}^{k-1}(2j-1)
$$
FINAL RESULT using the definition of double factorial (find it here): 
$$
f^{(k)}_{x=1}=\frac{(-1)^{k+1} 2k}{2^k\sqrt 2}(2k-3)!!
$$
$$
f^{(2015)}_{x=1}=\frac{4030(4027)!!}{2^{2015}\sqrt 2}\approx 1.03\times 10^{5783}
$$
Remark: the derivative is calculated in 0 for t, this means that it is calculated in $x=1$, just as you wanted.
