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Answer 1 · 2017-03-02T09:03:16

$\frac{d^{n}}{{(d x)}^{n}} (\frac{1}{x}) = {(- 1)}^{n} \frac{n!}{x^{n + 1}}$ $d^n/(dx)^n(1/x) = (-1)^n(n!)/x^(n+1)$

We have that:

$(1) f^{(n)} (x) = {(- 1)}^{n} (n!) x^{- (n + 1)}$ $(1) f^((n))(x) = (-1)^n(n!)x^(-(n+1))$

We can prove it directly for $n = 1$ $n=1$ :

$f'(x) = d/dx(1/x) = -1/x^2 = (-1)^1(1!)x^(-(1+1))$

In general, suppose:

$f^((n))(x) = (-1)^n(n!)x^(-(n+1))$

we have that:

$f^((n+1))(x) = d/dx f^((n))(x) = (-1)^n(n!)(-(n+1))x^(-(n+2)) = (-1)^(n+1)((n+1)!)x^(-(n+2))$

So the equation (1) is proven by induction.