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.

• Ok usually you try to find some dependence of the number of derivatives and your derivative and guess an answer which you show then by induction; I try to be precisely in an answer – Probabilitytheoryapprentice Aug 25 '16 at 9:30
• Here $f''(1)=0$ – iamvegan Aug 25 '16 at 9:33
• Not that it is necessary here, but ... Have you covered Taylor series? Thinking in terms of Taylor series makes this easy. Let $t=x-1$. In terms of that variable you are looking at the function $f(t)=t/\sqrt{2-t}$ and its derivatives at $t=0$. If you can find the $2014$th derivative of $1/\sqrt{2-t}$ at $t=0$, then Taylor series manipulations do the rest. – Jyrki Lahtonen Aug 25 '16 at 9:49
• @JyrkiLahtonen That could work well. Will try it as well later. – MathIsTheWayOfLife Aug 25 '16 at 9:56

Hint $$f(x)= \frac{x-1}{\sqrt{3-x}}=\frac{2}{\sqrt{3-x}}-\frac{3-x}{\sqrt{3-x}}=2(3-x)^{-\frac{1}{2}}-(3-x)^{\frac{1}{2}}$$

• oh you already posted. I was goint to put same hint. cheers! – Max Payne Aug 25 '16 at 9:35

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

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

• +1. Don't try to write down explicitly the whole answer (2). It's pretty long. – Felix Marin Aug 25 '16 at 21:39

Hint. Observe that $$\frac{x-1}{\sqrt{3-x}}=-\sqrt{3-x}+\frac{2}{\sqrt{3-x}}.$$ Then, show inductively that, for $n\ge 2$ $$\frac{d^n}{dx^n}\sqrt{3-x}=-\frac{(2n-3)!}{2^{2n-2}(n-2)!}(3-x)^{-\frac{2n-3}{2}}$$

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 $$f^{(2015)}(1)=((x-1)\cdot (3-x)^{-1/2})^{(2015)}\Big|_{x=1}={2015 \choose 1}(x-1)^{(1)}((3-x)^{-1/2})^{(2015-1)}\Big|_{x=1}\\ ={2015 \choose 1}(-1)^{2014}(3-x)^{-1/2-2014}4027!!/2^{2014}\Big|_{x=1}\\ =2015(-1)^{2014}4027!!2^{-(1/2+4028)}$$.

$$f(x)=\frac{x-1}{\sqrt{3-x}}$$ 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.