Prove all derivatives of $f(x)=\frac{1}{1+x}$ by induction  Problem 
Prove all derivatives of:
$$ f(x)=\frac{1}{1+x} $$
by induction.
 Attempt to solve 
I compute few derivatives of $f(x)$ so that i can form general expression for induction hypothesis. I compute all derivatives utilizing formula:
$$ \frac{d}{dx}x^n=nx^{n-1} $$
First 4 derivatives are:
$$ f'(x)=(-1)\cdot(1+x)^{-2}\cdot 1 = -\frac{1}{(1+x)^2} $$
$$ f''(x)=(-1)(-2)(1+x)^{-3}\cdot 1 = \frac{2}{(1+x)^3} $$
$$ f'''(x)=(-1)(-2)(-3)(1+x)^{-4}\cdot 1 = -\frac{6}{(1+x)^4} $$
$$ f''''(x)=(-1)(-2)(-3)(-4)(1+x)^{-5} \cdot 1 = \frac{24}{(1+x)^5} $$
Observe that $(-1)(-2)(-3)(-4)\dots (-n)$ can be generalized with:
$$ (-1)(-2)(-3)(-4)\dots(-n) = (-1)^n\cdot n! $$
Expression follows factorial of $n$ except every other value is positive and every other is negative. If i multiply it by $(-1)^n$ it is positive when $n \mod 2 = 0$ and negative when $n \mod 2 \neq 0$.
Rest of the expression can be generalized as:
$$ (1+x)^{-n-1} = (1+x)^{-(n+1)}=\frac{1}{(1+x)^{n+1}} $$
Combining these gives formula in analytic form:
$$ f(n)= \frac{(-1)^n \cdot n!}{(1+x)^{n+1}} $$
I can form induction hypothesis such that:
$$ \frac{d^n}{dx^n}\frac{1}{1+x} = \frac{(-1)^n\cdot n!}{(1+x)^{n+1}} $$
Induction proof
Base case
Base case when $n=0$:
$$ \frac{d^0}{dx^0}\frac{1}{1+x}=\frac{1}{1+x}=\frac{(-1)^0\cdot 0!}{(1+x)^{0+1}} $$
Induction step
$$ \frac{d^n}{dx^n}\frac{1}{1+x} =_{\text{ind.hyp}} \frac{(-1)^n\cdot n!}{(1+x)^{n+1+1}} $$
$$ \frac{d^n}{dx^n}\frac{1}{1+x} = \frac{(-1)^n\cdot n!}{(1+x)^{n+2}} $$
Now the problem is that formula i used for derivation can only be used recursively. I believe this is correct notation for $n$:th derivative but computing one is only defined recursively with formula i used:
$$ \frac{d}{dx} x^n = nx^{n-1} $$
Which is not defined for case:
$$ \frac{d^n}{dx^n}x^n = \text{
undefined} $$
The idea is to show that this recursion can be expressed in analytical form and it is valid for all $n\in \mathbb{Z}+$ by induction. Problem is i don't know how do you express this in recursive form and how do you get from recursion formula to the analytical one. 
 A: Your hypothesis should be
$$\frac{\mathrm d^n}{\mathrm dx^n}\frac{1}{1+x} = \frac{(-1)^n n!}{(1+x)^{n+1}}$$
and what you need for your induction step is:
$$\frac{\mathrm d^{n+1}}{\mathrm dx^{n+1}}f(x)= \frac{\mathrm d}{\mathrm dx}\left( \frac{\mathrm d^n}{\mathrm dx^n}f(x)\right) \quad \forall n \ge 0$$
A: Your hypothesis is 
$$\frac{d^n}{dx^n}\frac1{1+x}=\frac{(-1)^nn!}{(1+x)^{n+1}}$$
We want to show that 
$$\frac{d^{n+1}}{dx^{n+1}}\frac1{1+x}=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}}$$
Let's verify:
\begin{align}
\frac{d^{n+1}}{dx^{n+1}}\frac1{1+x} &=\frac{d}{dx}\left(\frac{d^n}{dx^n}\frac1{1+x} \right)\\
&=\frac{d}{dx}\left(\frac{(-1)^nn!}{(1+x)^{n+1}} \right) \\
&=(-1)^n(n!) \frac{d}{dx}(1+x)^{-(n+1)} \\
&= (-1)^n(n!) (-(n+1)) (1+x)^{-(n+2)}\\
&=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}}
\end{align}
A: You inductive hypothesis is that 
$$
\frac{d^nf}{dx^n}=\frac{(-1)^nn!}{(1+x)^{n+1}}.
$$
Your inductive step would then be
$$
\frac{d^{n+1}f}{dx^{n+1}}=\frac d{dx}\,\frac{d^nf}{dx^n}=\frac d{dx}\left(\frac{(-1)^nn!}{(1+x)^{n+1}}\right)
=\frac{-(n+1)(-1)^nn!}{(1+x)^{n+2}}=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}},
$$
which shows that the formula holds. 
