Question

How do you find the formula for the derivative of `1/x`?

Answer 1

`(d x^a)/(dx) = a*x^(a-1)`

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

Therefore
`(d 1/x)/(dx) = -x^(-2)` or `- 1/(x^2)`

Answer 2

I will assume that you are working from first principles, that is the definition.

There are two choices for how to "officially" define the derivative, every textbook author and teacher makes a decision which one to make official.

I'll use: `f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h`

For `f(x) = 1/x`, we get:

`f'(x) = lim_(h rarr0) (f(x+h)-f(x))/h`

`color(white)"ssssss"` `= lim_(h rarr0) (1/(x+h)-1/x)/h`

`color(white)"ssssss"` `= lim_(h rarr0) ((x-(x+h))/(x(x+h)))/(h/1)`

`color(white)"ssssss"` `= lim_(h rarr0) ((x-x-h))/(x(x+h))*(1/h)`

`color(white)"ssssss"` `= lim_(h rarr0) ((-1))/(x(x+h))`

`color(white)"ssssss"` `= -1/x^2`

That is: `f'(x) = -1/x^2`