Question

Question #6b550

Answer 1

By observing the first few derivatives,

`f(x)=x^{-1}`

`f'(x)=(-1)x^{-2}`

`f''(x)=(-1)(-2)x^{-3}`

`f'''(x)=(-1)(-2)(-3)x^{-4}`

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`f^{(n)}(x)=(-1)(-2)(-3)cdotcdotscdot(-n)x^{-(n+1)}`

Hence,

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

I hope that this was helpful.