Finding n-th derivative How can I find the n-th derivative of $\frac{1}{x(x+1)}$ ? I tried expressing it as such, but I'm not sure about it.
$\frac{1}{x(x+1)} = \frac{1}{x}-\frac{1}{x+1}$
edit: I think I found it, can anyone confirm?
Let $f_n$ be the n-th derivative of $f$
$f_1\left(x\right)=-\frac{1}{x^2}+\frac{1}{\left(x+1\right)^2}$
$f_2\left(x\right)=\frac{2}{x^{\text{3}}}-\frac{2}{\left(x+1\right)^3}$
$f_3\left(x\right)=-\frac{6}{x^4}+\frac{6}{\left(x+1\right)^4}$
$f_4\left(x\right)=\frac{24}{x^5}-\frac{24}{\left(x+1\right)^5}$
We can then guess that the n-th derivative would be:
$f_n\left(x\right)=\left(-1\right)^n\left(-\frac{n!}{x^{n+1}}+\frac{n!}{\left(x+1\right)^{n+1}}\right)$
However is this enough to "proof" the result? Like is it acceptable if I was in, say, an exam? $f_n$ is simply based on an assumption?...
Also, how can I find x such as
$f_n(x) = 0$ ?
 A: First of all: your derivatives and the guessed pattern are nearly (fix the exponent of the $(-1)$) correct, good job!
"However is this enough to "proof" the result?" Not really. (About if this is acceptable in an exam, I can't possibly know, it depends on the exam.) You only guessed the pattern and you cannot, in general, be sure from the first few results, that this pattern continues, so this does not qualify as a (formal) proof.
Can you prove your result by induction, which is the standard approach here? Think about what the base case would look like and for the induction step show that the derivative of your proposed $n$-th derivative is in fact your proposed $(n+1)$-th derivative?
For your second question: if you set the derivative to zero, first cancel out all the constant terms. What do you end up with?
A: Pattern-spotting doesn't rise to a proof by induction, but it gives you the theorem such a proof needs to prove, viz.$$\frac{d^n}{dx^n}\frac{1}{x+c}=\frac{(-1)^nn!}{(x+c)^{n+1}}.$$A proof by induction begins with a trivial $n=0$ base step, viz. $\frac{1}{x+c}=\frac{1}{x+c}$; for the inductive step,$$\frac{d^k}{dx^k}\frac{1}{x+c}=\frac{(-1)^kk!}{(x+c)^{k+1}}\implies\frac{d^{k+1}}{dx^{k+1}}\frac{1}{x+c}=\frac{d}{dx}\frac{(-1)^kk!}{(x+c)^{k+1}}=\frac{(-1)^{k+1}(k+1)!}{(x+c)^{k+2}}.$$The cases $c\in\{0,\,1\}$ allow you to complete the proof, using $\frac{d^n}{dx^n}(f-g)=\frac{d^nf}{dx^n}-\frac{d^ng}{dx^n}$, which is itself provable by induction on $n$.
