Answer to Question #258552 in Calculus for Fiona

Question

f^(7) (x) of y = x^-2

Answer 1

Let $P(n)$ be the proposition that $f^{(n)}=(-1)^{n}(n+1)!x^{-(n+2)}, n\geq1$

Basis Step

$P(1)$ is true, because

$$f'(x)=(x^{-2})'=-2x^{-3}=(-1)^1(1+1)!x^{-(1+2)}$$

Inductive Step

Assume that $P(k)$ holds for an arbitrary integer $k\geq1.$ That is, we assume that

$$f^{(k)}=(-1)^{k}(k+1)!x^{-(k+2)}, k\geq1$$

Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that

$$f^{(k+1)}=(-1)^{k+1+1}(k+1)!x^{-(k+1+2)},$$

is also true.

$$f^{(k+1)}=(f^{(k)})'=((-1)^{k}(k+1)!x^{-(k+2)})'$$

$$=(-1)^{k}(k+1)!(-(k+2))x^{-(k+2)-1}$$

$$=(-1)^{k+1}(k+2)!x^{-(k+1+2)}$$

$$=(-1)^{k+1}(k+1+1)!x^{-(k+1+2)}$$

This last equation shows that $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we

know that $P(n)$ is true for all integers $n\geq1.$ That is, we have proven that 

$$f^{(n)}=(-1)^{n}(n+1)!x^{-(n+2)}, n\geq1$$

Then

$$f^{(7)}=(-1)^{7}(7+1)!x^{-(7+2)}=-8!x^{-9}$$

$$=-40320x^{-9}$$