Spotting a pattern for the $n-th$ derivative I have a function $y=\sqrt{\frac {x+1}{1-x}}$ and I am trying to find a pattern for the $n-th$ derivative, I have so far found this:
$y'={1\over (x-1)^2({\frac {x+1}{1-x}})^{1/2}}$
$y''={2x+1\over (x-1)^4{(\frac {x+1}{1-x})}^{3/2}}$
$y'''= {3(2x^2+2x+1)\over (x-1)^6{(\frac {x+1}{1-x})}^{5/2}}$
I can see a pattern forming on the denominator but I can't seem to find a pattern on the numerator that I can express in terms of $n$. Any ideas?
 A: $y(x)=\exp\operatorname{arctanh}(x)$ leads to $y'(x)=\frac{y(x)}{1-x^2}$ and to 
$$ (1-x^2) y'(x) = y(x)\tag{1} $$
where by differentiating both sides and rearranging we get
$$ (1-x^2)y''(x) = (1+2x)y'(x) \tag{2} $$
and by differentiating again we get
$$ (1-x^2)y'''(x) = (1+4x) y''(x)+ 2 y'(x)\tag{3} $$
$$ (1-x^2)y^{(4)}(x) = (1+6x)y'''(x)+6y''(x)\tag{4} $$
$$ (1-x^2)y^{(5)}(x) = (1+8x)y^{(4)}(x)+12y'''(x)\tag{5} $$
so by induction
$$ (1-x^2) y^{(n+1)}(x) = (1+2nx) y^{(n)}(x)+n(n-1) y^{(n-1)}(x) \tag{6}$$
which allows a recursive determination of $y^{(n)}(x)$.

Small addendum: of course, this is an unnecessary complication if you are just interested in the derivatives at the origin, since by Cauchy integral formula and conformal maps:
$$ y^{(n)}(0) = \frac{n!}{2\pi i}\oint_{\|z\|=\varepsilon}\frac{\exp\operatorname{arctanh}(z)}{z^{n+1}}\,dz = \frac{n!}{2\pi i}\oint_{\|z\|=\varepsilon}\frac{\exp(z)}{\tanh^{n+1}(z)\cosh^2(z)}\,dz$$
equals
$$ \frac{n!(-1)^{n+1}}{2\pi i}\oint_{\|u-1\|=\varepsilon}\frac{4u^2(1+u^2)^{n-1}}{(1-u^2)^{n+1}}\,du =n!(-1)^{n+1}\operatorname*{Res}_{u=1}\frac{4u^2(1+u^2)^{n-1}}{(1-u^2)^{n+1}}$$
or
$$n!(-1)^{n+1}\operatorname*{Res}_{u=1}\frac{2\sqrt{u}(1+u)^{n-1}}{(1-u)^{n+1}}=n!(-1)^n [v^n]2\sqrt{1-v}(2-v)^{n-1}$$
or
$$ n!\cdot \sum_{k=0}^{n-1}\binom{n-1}{k}\binom{\frac{1}{2}}{n-k}2^{n-k}. $$
