Solutions

a) Let $g(x)=x-a$ and $h(x)=\frac{1}{x^n}$. Then we can write

$$f(x)=h(g(x))=\frac{1}{(x-a)^n}.$$

By the chain rule,

$$f^{(1)}(x)=h'(g(x))g'(x)=h'(g(x)).$$

It follows that higher-order derivatives respect the equation

$$f^{(k)}(x)=h^{(k)}(g(x))=(-1)^k\frac{(n+k-1)!}{(n-1)!}(x-a)^{-n-k},$$

with the rightmost expression a result of Exercise 30.

b) Notice that

$$f(x)=\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right).$$

We therefore have by part a)

$$\begin{align*} f^{(k)}(x)&=\frac{1}{2}\left(\frac{(-1)^kk!}{(x-1)^{k+1}}-\frac{(-1)^kk!}{(x+1)^{k+1}}\right) \\ &=\frac{1}{2}(-1)^kk!\left(\frac{1}{(x-1)^{k+1}}-\frac{1}{(x+1)^{k+1}}\right). \end{align*}$$