Question

How do you use the definition of a derivative to show that if `f(x)=1/x` then `f'(x)=-1/x^2`?

Answer 1

This is a proof

Explanation:

By definition:
`f'(x) = lim_(h->0)(f(x+h)-f(x))/h`

so, with `f(x)=1/x` we have:

`f'(x) = lim_(h->0)((1/(x+h)-1/x))/h`

`= lim_(h->0)((x-(x-h))/(x(x+h)))/h`

`= lim_(h->0)((x-x-h)/(x(x+h)))/h`

`= lim_(h->0)((-h)/(x(x+h)))/h`

`= lim_(h->0)(-1)/(x(x+h))`

`= lim_(h->0)(-1)/(x^2+xh)`

`= -1/(x^2)` QED