Question

What is the derivative of `x^-1`?

Answer 1

You get: `-1/x^2`

Explanation:

You can use:
1] The Power Rule where the derivative of `x^n` is `nx^(n-1)` and get:
`y'=-1x^-2=-1/x^2`

2] The definition of derivative as:
`y'=lim_(h->0)(f(x+h)-f(x))/h`
so you get:
`y'=lim_(h->0)((x+h)^-1-(x)^-1)/h=`
but `x^-1=1/x`
so you get:
`y'=lim_(h->0)(1/(x+h)-1/(x))/h` rearranging:
`y'=lim_(h->0)[(x-x-h)/(x(x+h))]*1/h=`
`y'=lim_(h->0)[(cancel(x)cancel(-x)-cancel(h))/(x(x+h))]*1/cancel(h)=`
taking the limit:
`y'=-1/x^2`