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!
 A: You can maybe try to examine the result for the first values of $n$ and then try an induction on $n$.
Let $f$ be the function $\displaystyle x\mapsto\frac{7}{1-x}$, one has: $$f'(x)=\frac{7}{(1-x)^2},f''(x)=\frac{14}{(1-x)^3},f'''(x)=\frac{42}{(1-x)^4}.$$
One may conjecture that: $$\forall n\in\mathbb{N},f^{(n)}(x)=\frac{7(n!)}{(1-x)^{n+1}}.$$
You can prove this statement using induction on $n$ and the formula to differentiate $1/u$.
A: Claim: $$f^{(n)}(x) = \frac{7(n!)}{(1-x)^{n+1}}$$ where $f^{(n)}(x)$ denotes the $n$-th derivative of $y = f(x)$.
So, assuming $f^{(k)} = \frac{7(k!)}{(1-x)^{k+1}}$, differentiate this to get: 
$$f^{(k+1)} = 7(k!)(k+1) \cdot \frac{1}{(1-x)^{k+2}} = \frac{7(k+1)!}{(1-x)^{(k+1) + 1}}$$
as required. So our claim holds for all natural $n$.
This allows us to write down the McLaurin expansion of $f(x)$, since $$f(x) = f(0) + f'(0)x + \frac{1}{2!}f''(0)x^2 + \cdots + \frac{1}{n!}f^{(n)}(0)x^n + \cdots$$
But we've just computed $f^{(n)}(x)$, so finding $f^{(n)}(0)$ follows easily. 
(alternatively, you could have written down the Taylor series by noticing that $y$ is a geometric series)
A: 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 $$f(c)= \frac{7(n!)}{(1-c)^{n+1}}\, \forall c \in (\mathbb{R}-\{1\})$$
