Trick of doing successive differentiation I noticed that in most of example of successive had done by following few steps. At first they made a function in denominator something just like this $$\frac{x}{(x+1)(x-1)}$$. In every examples they had followed the above step. Then, they had differentiate $n$ times. I think there's some symmetry. But, I can't understand cause, they didn't complete whole sum. When I differentiate $1$ times. I get an answer which is bigger than as I expect. Is there actually any trick to solve successive differentiation easy way?
 A: I am not sure if I understand you correctly. Usually the terms will be more complicated after each differentiation, especially if the product or the  composition of functions is involved. In some cases these products and compositions can be transformed into a simple structured expressions that can be easily be differentiated mutltiple time. This is one of these terms, because
$$\frac{x}{(x+1)(x-1)}=\frac 1 2 \left(\frac 1 {x+1} + \frac 1  {x-1}\right)=\frac 1 2(x+1)^{-1}+\frac 1 2(x-1)^{-1}$$
The right hand side can be easiliy differentiated
$$\left(\frac 1 2(x+1)^{-1}+\frac 1 2(x-1)^{-1}\right)^{(n)}=$$
$$\frac {(-1)^n n!}2(x+1)^{-n}+\frac {(-1)^n n!}2(x-1)^{-n}$$
A: Some tips:

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*Partial fractions. $\frac {x}{(x+1)(x-1)}=$ $x\left(\frac {1/2}{x-1}-\frac {1/2}{x+1}\right)=$ $\frac {x/2}{x-1}-\frac {x/2}{x+1}=$ $(1/2)(\,(1+\frac {1}{x-1})-(1-\frac {1}{x+1})\,)=$ $(1/2)(\frac {1}{x-1}+\frac {1}{x+1}).$


*Generalized product rules.(i). Let $u^{(0)}=u$ and let $u^{(j)}$ be the $j$th derivative of $u$ when $j>0.$ The generalization of the formula $(uv)'=uv'+u'v$ is $$(uv)^{(n)}=\sum_{j=0}^n\binom {n}{j}u^{(j)}v^{(n-j)}.$$ (ii). For the $1$st derivative of a product $P=\prod_{i=1}^nu_i$ of non-$0$ functions, we have $$P'=(P)\sum_{i=1}^n\frac {u_i'}{u_i}.$$ If you simplify each term $(P)\frac {u_i'}{u_i} $you get a valid formula for $P'$ regardless of the value any $u_i.$ E.g. $$(abc)'=a'bc+ab'c+abc'.$$
