# 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.

• All of your work seems fine. I'm not sure why the formula for iterated derivatives of $x^n$ is causing you anxiety, as you're only using it for negative values of $n$ (thus you'll never be in a position of needing the value of $\frac{d^n}{dx^n} x^n$). – Connor Harris Oct 4 '18 at 16:40
• I am not sure why you say that $\frac{d^n}{dx^n}x^n$ is undefined. Simply applying it to a couple of values from $\mathbb{Z}^+$ leads to the intuitive expression $\frac{d^n}{dx^n}x^n=n!$ – mrtaurho Oct 4 '18 at 16:40

$$\frac{d^n}{dx^n}\frac1{1+x}=\frac{(-1)^nn!}{(1+x)^{n+1}}$$

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}

$$\frac{\mathrm d^n}{\mathrm dx^n}\frac{1}{1+x} = \frac{(-1)^n n!}{(1+x)^{n+1}}$$
$$\frac{\mathrm d^{n+1}}{\mathrm dx^{n+1}}f(x)= \frac{\mathrm d}{\mathrm dx}\left( \frac{\mathrm d^n}{\mathrm dx^n}f(x)\right) \quad \forall n \ge 0$$
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.