For example, If I had to calculate the first, second and third derivative of $f(x)=\frac{1}{x-1}$
I would find that they would be the following:
$$f^{'}(x)=-\frac{1}{\left(x-1\right)^2}$$ $$f^{''}(x)=\frac{2}{\left(x-1\right)^3}$$ $$f^{3}(x)=-\frac{6}{\left(x-1\right)^4}$$
Seeing this pattern, I want to understand what $f^{\left(n\right)}\left(x\right)\:-the\:n^{th}\:derivative$ means in relation to this pattern.
On
yes $$f^{(n)}(x)$$ means the n-th derivative. i think it must be $$f^{(n)}(x)=(-1)^{(n)}n!(x-1)^{-(n+1)}$$
On
Hint. Assume $\alpha \in \mathbb{R}$ and $\alpha \ne 0$. One may prove by a direct recurrence $$ \left(u^\alpha\right)^{(k)}:=\frac{d^k}{du^k}\left(u^\alpha\right)=\alpha (\alpha-1)\cdots (\alpha-k+1)\cdot u^{\alpha-k}, \qquad k=0,1, \cdots. $$ By putting $$ u=x-1,\quad \alpha=-1, \quad k=n, $$ one gets here
$$ f^{(n)}(x)= (-1)^{n} \cdot\frac{n!}{(x-1)^{n+1}} $$
with $f(x)= \dfrac{1}{(x-1)}$.
The nth derivative has the same meaning as, say, the fourth derivative, except that in this case $n $ could be any arbitrary natural number. To find an nth derivative you need to observe a pattern in the derivatives, and if you're feeling cautious of your choice, prove it by induction.
Here, we have an oscillating negative sign, where the negatives show up on the odd terms. This implies the existence of $(-1)^n$ in the nth derivative.
Carrying on in a similar fashion, I obtained $$f^{(n)}(x) = (-1)^n \frac{n!}{(x-1)^{n+1}}$$
Do you see why this answer works?
By all means, prove it!