Answer
a) $f^{(n)}(x)=n!$
b) $f^{(n)}(x)=\frac{(-1)^nn!}{x^{n+1}}$
Work Step by Step
Part a)
$f(x)=x^n$
$f'(x)=nx^{n-1}$
$f''(x)=n(n-1)x^{n-2}$
$f'''(x)=n(n-1)(n-2)x^{n-3}$
Continuing this process, we get
$f^{(n-1)}(x)=n(n-1)(n-2)\ldots (n-(n-2))x^{n-(n-1)}=n(n-1)(n-2)\ldots 2x$
$f^{(n)}(x)=n(n-1)(n-2)\ldots 2\cdot 1=n!$
Thus, $f^{(n)}(x)=n!$
Part b)
$f(x)=1/x=x^{-1}$
$f'(x)=-1x^{-2}$
$f''(x)=1\cdot 2x^{-3}=2!x^{-3}$
$f'''(x)=-1\cdot 2\cdot 3 x^{-4}=-3!x^{-4}$
$f^{(4)}(x)=1\cdot 2\cdot 3\cdot 4x^{-5}=4!x^{-5}$
Continuing this process, we get
$f^{(n)}=(-1)^nn!x^{-(n+1)}=\frac{(-1)^nn!}{x^{n+1}}$
Thus, $f^{(n)}(x)=\frac{(-1)^nn!}{x^{n+1}}$.
