Question

How do you find f'(x) using the limit definition given `f(x)=3x^(−2)`?

Answer 1

`= - 6/x^3`

Explanation:

by definition

`f'(x) =lim_{h to 0} (f(x+h) - f(x))/(h)`

`=lim_{h to 0} (1/h) (3/(x+h)^2 - 3/x^2)`

`=lim_{h to 0} (1/h)* (3x^2 - 3(x+h)^2)/((x+h)^2x^2)`

`=lim_{h to 0} (1/h) * (3x^2 - 3(x^2+2xh+h^2))/((x+h)^2x^2)`

`=lim_{h to 0} (1/h)* ( - 6xh-3h^2)/((x+h)^2x^2)`

`=lim_{h to 0}( - 6x-3h)/((x+h)^2x^2)`

`=lim_{h to 0}( - 6x)/((x+h)^2x^2) + mathcal(O)(h)`

`=-( 6x)/((x+0)^2x^2)`

`=-( 6x)/x^4`

`= - 6/x^3`

Answer 2

I found `f'(x)=-6/x^3`

Explanation:

Using the limit definition we get:
`f'(x)=lim_(h->0)(f(x+h)-f(x))/h`
where `h` is a small increment.
In our case we have (using the fact that `x^-2=1/x^2`):
`f'(x)=lim_(h->0)(3/(x+h)^2-3/(x^2))/h=`
`=lim_(h->0)1/h((3x^2-3(x+h)^2)/((x+h)^2x^2))=`
`=lim_(h->0)1/h((cancel(3x^2)cancel(-3x^2)-6xh-3h^2)/((x+h)^2x^2))=`
`=lim_(h->0)1/cancel(h)(cancel(h)(-6x-3h)/((x+h)^2x^2))=`
as `h->0` we get:
`=lim_(h->0)((-6xcancel(-3h)^0)/((x+cancel(h)^0)^2x^2))=(-6x)/x^4=-6/x^3`