Answer
$(a) f^{(n)}(x)=n!$
$(b)f^{(n)}(x)=(-1)^n\frac{n!}{x^{n+1}}$
Work Step by Step
$(a)$ $f(x)=x^n$. Differentiating with respect to $x$, $$f'(x)=nx^{n-1}$$Differentiating again, $$f''(x)=n(n-1)x^{n-2}$$Differentiating again, $$f'''(x)=n(n-1)(n-2)x^{n-3}$$From here we can see the pattern and suggest that, $$f^{(n)}(x)=n(n-1)(n-2)...x^{n-n}=n!x^0=n!$$
$(b)$ $f(x)=\frac1x=x^{-1}$. Differentiating with respect to $x$, $$f'(x)=-1\times x^{-2}=-x^{-2}$$ Differentiating again, $$f''(x)=(-1)(-2)x^{-3}=2x^{-3}$$Differentiating again, $$f'''(x)=(2)(-3)x^{-4}=-6x^{-4}$$From here we can observe the pattern, $$f^{(n)}=(-1)^nn!x^{-(n+1)}=(-1)^n\frac{n!}{x^{n+1}}$$
