Find the \(n^{th}\) derivative of \(y=\dfrac{x+1}{x^2-4}\).

Answer: \(y_n=(-1)^n\,n!\left[\dfrac{3}{4(x-2)^{n+1}}+\dfrac{1}{4(x+2)^{n+1}}\right]\).

Partial fractions: \(\dfrac{x+1}{(x-2)(x+2)}=\dfrac{A}{x-2}+\dfrac{B}{x+2}\).

Put \(x=2\): \(3=4A\Rightarrow A=\frac34\). Put \(x=-2\): \(-1=-4B\Rightarrow B=\frac14\).

\[y=\frac34\cdot\frac{1}{x-2}+\frac14\cdot\frac1{x+2}.\]

Standard result: \(\dfrac{d^n}{dx^n}\left(\dfrac{1}{ax+b}\right)=\dfrac{(-1)^n\,n!\,a^n}{(ax+b)^{n+1}}\). Here \(a=1\).

\[y_n=\frac34\cdot\frac{(-1)^n n!}{(x-2)^{n+1}}+\frac14\cdot\frac{(-1)^n n!}{(x+2)^{n+1}}=(-1)^n n!\left[\frac{3}{4(x-2)^{n+1}}+\frac{1}{4(x+2)^{n+1}}\right].\]

Check \(n=1\): \(y_1=-\frac{3}{4(x-2)^2}-\frac1{4(x+2)^2}\); at \(x=0\) this gives \(-\frac{3}{16}-\frac1{16}=-\frac14\), and direct differentiation \(y'=\frac{(x^2-4)-2x(x+1)}{(x^2-4)^2}\) at \(x=0\) gives \(\frac{-4}{16}=-\frac14\). ✓