Question

How do you use the definition of a derivative to find the derivative of `f(x)=(1/x)-2`?

Answer 1

I found `f'(x)=-1/x^2`

Explanation:

We use our definition as:
`f'(x)=lim_(h->0)(f(x+h)-f(x))/h`
where `h` is a small increment.

In our case:
`f'(x)=lim_(h->0)[(1/(x+h)-2)-(1/x-2)]/h=`
`=lim_(h->0)[(1/(x+h)cancel(-2)-1/xcancel(+2))]/h=`
`=lim_(h->0)(1/h((cancel(x)cancel(-x)-h)/(x(x+h)))=lim_(h->0)(1/cancel(h)(-cancel(h)/(x(x+h)))=lim_(h->0)(-1/(x(x+h)))=`
as `h->0`
`=(-1/(x(x+cancel(h))))=-1/x^2`