Answer
(a)$$f^n(x)=n!$$
(b)$$f^n(x)=\frac{(-1)^nn!}{x^{n+1}}$$
Work Step by Step
(a) $$f(x)=x^n$$
- First derivative : $f'(x)=\frac{d}{dx}(x^n)=nx^{n-1}$
- Second derivative : $f''(x)=n\frac{d}{dx}(x^{n-1})=n(n-1)x^{n-2}$
- Third derivative : $f'''(x)=n(n-1)\frac{d}{dx}(x^{n-2})=n(n-1)(n-2)x^{n-3}$
Observing the pattern, we predict the nth derivative would be $$f^n(x)=n\times(n-1)\times...\times[n-(n-3)]\times[n-(n-2)]\times[n-(n-1)]x^{n-n}$$
$$f^n(x)=n\times(n-1)\times...\times3\times2\times1\times x^0$$
$$f^n(x)=n\times(n-1)\times...\times3\times2\times1$$
$$f^n(x)=n!$$
(b) $$f(x)=\frac{1}{x}=x^{-1}$$
- First derivative: $f'(x)=\frac{d}{dx}(x^{-1})=-x^{-2}$
- Second derivative: $f''(x)=-\frac{d}{dx}(x^{-2})=-(-2)x^{-3}=2x^{-3}$
- Third derivative: $f'''(x)=2\frac{d}{dx}x^{-3}=2\times(-3)x^{-4}$
Observing the pattern, we predict the nth derivative would be $$f^n(x)=(-1)\times(-2)\times(-3)\times...\times(-n)x^{-(n+1)}$$
$$f^n(x)=\frac{(-1)\times(-2)\times(-3)\times...\times(-n)}{x^{n+1}}$$
$$f^n(x)=\frac{(-1)^n(1\times2\times3\times...\times n)}{x^{n+1}}$$
$$f^n(x)=\frac{(-1)^nn!}{x^{n+1}}$$
