Chapter 3, Problem 66RE

To determine

To prove: The $n\text{th}$ derivative of $x^n$ is $n\mathrm{<mo>!</mo>}$ for any positive integer $n$.

Answer to Problem 66RE

  $f^n\left(x\right)=n\left(n-1\right)\left(n-2\right)\mathrm{...2}\left(1\right)x^{n-n}=n!\left(1\right)=n!$

The above equation proves that the $n\text{th}$ derivative of $x^n$ is $n\mathrm{<mo>!</mo>}$ for any positive integer $n$.

Explanation of Solution

Given information:

  $n$ is a positive integer.

Let’s find the first four derivatives of $f\left(x\right)=x^n$

  $\begin{array}{c}{f}^{\prime }=n^{n-1}\\ f^{″}=n\left(n-1\right)x^{n-2}\\ f^{‴}=n\left(n-1\right)\left(n-2\right)x^{n-3}\\ f^{\left(4\right)}=n\left(n-1\right)\left(n-2\right)\left(n-4\right)x^{n-4}\end{array}$

From the above four derivatives of $x^n$ we can see that the power of $x$ is $n-k$ for each derivative $f^{\left(k\right)}\left(x\right)$ .

The last factor is $n-\left(k-1\right)$ . The $n\text{th derivative would have k=n }{x}^{n-k}=x^{n-n}\text{and the last factor of }$

  $n-\left(k-1\right)=n\left(n-1\right)=n-n+1=1$.

Therefore we can conclude that $f^n\left(x\right)=n\left(n-1\right)\left(n-2\right)\mathrm{...2}\left(1\right)x^{n-n}=n!\left(1\right)=n!$ has been proved.