Formula for higher order derivatives of rational functions I've encountered this question : Let $f(x) = 
\frac{x}{(x − 1)}$. Find $f′(x)$, $f′′(x)$, and a formula for $f^{(n)}(x)$.
I can compute the derivatives easily, but I don't see what the formula might be. Could anyone help please?
Thanks!
 A: You have $$f(x)=\frac{x}{x-1}=\frac{x-1+1}{x-1}=1+\frac{1}{x-1}$$ Hence for $n \ge 1$: $$f^{(n)}(x)=\frac{(-1) ^n n!}{(x-1)^{n+1}}$$
A: First, we solve the first derivatives manually:
$f(x)=\frac{x}{x-1}$
$f'(x)=-\frac{1}{(x-1)^2}$
$f''(x)=2\frac{1}{(x-1)^3}$
$f'''(x)=-6\frac{1}{(x-1)^4}$
By doing so, we see some pattern emerging for $n>0$. We guess that
$f^{(n)}(x)=(-1)^nn!\frac{1}{(x-1)^{n+1}}$
To show that this is true, we use induction:
1) the formula holds for n=1.
2) Assume the formula holds for n>0. Show that it holds for n+1:
\begin{align*}
f^{(n+1)}(x)&=\frac{d}{dx}f^{(n)}(x)\\
&=\frac{d}{dx}\left((-1)^nn!\frac{1}{(x-1)^{n+1}}\right)\\
&=(-1)^nn!\frac{d}{dx}\frac{1}{(x-1)^{n+1}}\\
&=(-1)^nn!(-1)(n+1)\frac{1}{(x-1)^{n+2}}\\
&=(-1)^{n+1}(n+1)!\frac{1}{(x-1)^{n+2}}
\end{align*}
This fits the formula again, so we are done.
A: Note that $$\dfrac{d}{dx}\ln f(x)=\dfrac{f'(x)}{f(x)}$$ 
and $$\dfrac{d^n}{dx^n}\left(\dfrac{1}{x-a}\right)=\dfrac{(-1)^nn!}{(x-a)^{n+1}}.$$ 
If $f(x)=\dfrac{(x-a_1)(x-a_2)\cdots(x-a_n)}{(x-b_1)(x-b_2)\cdots(x-b_n)},$ then $$\ln f(x)=\Big(\ln(x-a_1)+\cdots+\ln(x-a_n)\Big)-\Big(\ln(x-b_1)+\cdots+\ln(x-b_n)\Big).$$ 
Now differentiate $\ln f(x).$
