Prove by Induction Summation

Question

Prove by induction:

Given that $f(x) = x^{-1}$, then the $k$-th derivative of $f$ is given by $f^{\langle k\rangle}(x) = (−1)^k \cdot k!\;x^{−(k+1)}$ for all $k ≥ 1$.

How do I go about proving this? I am very confused as to what the problem is asking and how I can even use induction to solve it.

Thank you!

Answer 1

Your premise $P(k)$ is $f^{\langle k\rangle}(x) = (-1)^k\;k!\;x^{-(k+1)}$ given $f(x) = x^{-1}$

For the base case we need to show the truth of $P(0)$, that is, $f^{\langle 0\rangle}(x) = (-1)^0\; 0!\; x^{-1}$ .

For the inductive step. If you assume $P(k)$ can you prove $P(k+1)$ ?

Hint: $f^{\langle k+1\rangle}(x) = \frac {\mathrm d\;}{\mathrm d x}f^{\langle k\rangle}(x)$

Answer 2

Hint. If by hypothesis, for some non negative integer $k$, you have$$f^{\langle k\rangle}(x) = (−1)^k \cdot k!\;x^{−(k+1)}$$ then you may write $$f^{\langle k+1\rangle}(x) = (−1)^k \cdot k!\;\frac{d}{dx}x^{−(k+1)}=(−1)^k \cdot k!\;(−(k+1))x^{−(k+1)-1}=\ldots$$ Hoping you can take it from here.