Find the $N^{th}$ derivative of $$f(x) = \sqrt{\frac {1-x}{1+x}}$$
I have got $1^{st}$ derivative as: $\frac{-1}{(1-x)^{1/2}(1+x)^{3/2}}$
and $2^{nd}$ derivative as: $\frac{1-2x}{(1-x)^{3/2}(1+x)^{5/2}}$
and $3^{rd}$ derivative as: $\frac{-6x^2+6x-3}{(1-x)^{5/2}(1+x)^{7/2}}$
I can see how will the denominator gets it form but can you help me with the numerator?
Thanks. :)
Let be $\varphi(x)=\sqrt{\frac{x-1}{x+1}}=f(g(x))$ where $f(x)=\sqrt x$ and $g(x)=\frac{x-1}{x+1}=\frac{2}{x+1}-1$.
It's easy to see that $$ f^{(n)}(x)=\frac{\sqrt\pi x^{\frac{1}{2}-n}}{2\Gamma\left(\frac{3}{2}-n\right)} $$ amd $$ g^{(n)}(x)=(-1)^n2 \,n!\,(x+1)^{-1-n} . $$
Thus, we can use the Faà di Bruno's formula in the classical form $$ \varphi^{(n)}(x)=\frac{d^n}{dx^n} f(g(x)) =\sum \frac{n!}{m_1!\,m_2!\,\cdots\,m_n!}\cdot f^{(m_1+\cdots+m_n)}(g(x))\cdot \prod_{j=1}^n\left(\frac{g^{(j)}(x)}{j!}\right)^{m_j} $$ where the sum is over all $n$-tuples of nonnegative integers $(m_1,\ldots,m_n)$ satisfying the constraint $1\cdot m_1+2\cdot m_2+3\cdot m_3+\cdots+n\cdot m_n=n$ or the Faà di Bruno's formula expressed in terms of Bell polynomials $B_{n,k}(x_1,\ldots,x_{n−k+1})$ $$\varphi^{(n)}(x)=\frac{d^n}{dx^n} f(g(x))= \sum_{k=1}^n f^{(k)}(g(x))\cdot B_{n,k}\left(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\right).$$
