Compute $\frac{d^ny}{dx^n}$ if $y = \frac{7}{1-x}$ I am wondering what is $\frac{d^n}{dx^n}$ if $y = \frac{7}{1-x}$
Basically, I understand that this asks for a formula to calculate any derivative of f(x) (correct me if I'm wrong). Is that related to Talylor's theory? How do I end up with such a formula? 
Thanks!

Let $f(x)= \sum_{n\geq 0} a_n(x-c)^n$
Then $f^{(n)}(c)=n!a_n  $, 
\begin{align*}
\frac{7}{1-x} &=    \frac{7}{1-c-(x-c)} \\
&= \frac{7}{1-c}\frac{1}{1-\frac{x-c}{1-c}}\\
&= \frac{7}{1-c}\sum_{n\geq 0    } \frac{1}{(1-c)^n} (x-c)^n\\
&=\sum_{n\geq 0    } \frac{7}{(1-c)^{n+1}} (x-c)^n
\end{align*}
Therefore $$f(c)= \frac{7(n!)}{(1-c)^{n+1}}\, \forall c \in (\mathbb{R}-\{1\})$$
