Nth Derivative of x^(n-1)

Question

how can i demonstrate this ? :

$$\ \frac{d^n[x^{n-1}]}{dx^n} =0 \quad, \forall n\in\mathbb N*$$

i started like this :

$$\ f(x)=x^{n-1} $$

$$\ f'(x)=(n-1)x^{n-2} $$

$$\ f''(x)=(n-2)(n-1)x^{n-3} $$

$$\ f^n(x)= \, ? \, $$

maybe something like $$\ (n-1)!x^{(n-1)-n} $$ but $$\ (n-1) < n $$

Thanks a lot.

Answer 1

The following is a proof by induction.

Claim : $\dfrac{d^k}{dx^k} (x^n) = \frac{n!}{(n-k)!}x^{n-k} , 0 \leq k \leq n$

Proof:

The base case is $k=0$, for which the function is $x^n$.

Suppose that we are given that $\dfrac{d^m}{dx^m} (x^n) = \frac{n!}{(n-m)!}x^{n-m}$. Then, $$ \dfrac{d^{m+1}}{dx^{m+1}} (x^n) =\dfrac{d}{dx}\dfrac{d^{m}}{dx^{m}} (x^n) =\frac{n!}{(n-m)!}(n-m)x^{n-m-1} =\frac{n!}{(n-m-1)!}x^{n-m-1} $$

Hence, the induction is complete.

Now, put $k=n$, and you get $\dfrac{d^n}{dx^n} (x^n) = n!$, which is a constant. Hence, $\dfrac{d^{n+1}}{dx^{n+1}} (x^n) = 0$. Finally, you can replace $n$ with $n-1$ everywhere to complete the proof.

Answer 2

Base Case: $$ (c)'=0 $$ For some constant $c$, a degree 0 polynomial.

Inductive step: Suppose $$ \frac{d^n}{dx^n}x^{n-1}=0 $$ then $$ \frac{d^{n+1}}{dx^{n+1}}x^{n}=\frac{d^{n+1}}{dx^{n+1}}x*x^{n-1}\\ =[\frac{d^{n+1}}{dx^{n+1}}x]*x^{n-1}+[\frac{d^{n+1}}{dx^{n+1}}x^{n-1}]*x $$ by the product rule. But the left term is certainly zero and the right hand side is zero by the inductive hypothesis.

Answer 3

Using Leibniz's formula $(fg)^{(n)} =\sum_{k=0}^n \binom{n}{k} f^{(k)}g^{(n-k)} $.

Set $f(x) = x$.

Since $f'(x) = 1$ and $f^{(m)}(x) = 0$ for $m \ge 2$, $(xg)^{(n)} =x g^{(n)}(x)+g^{(n-1)}(x) $.

Now set $g(x) = x^{n-2} $. This becomes

$(x^{n-1})^{(n)} =x (x^{n-2})^{(n)}(x)+(x^{n-2})^{(n-1)}(x) $.

If (this is our induction hypothesis, which is true for $n=2$ and $n=3$) $(x^{n-2})^{(n-1)}(x) =0$, then, also $(x^{n-2})^{(n)}(x) =0$ so that $(x^{n-1})^{(n)} =0$.