calculus
posted by help needed on .
f(x)=1/x^2
use formal definition of derivative to find slope at x=2

f'(x) = lim(h>0) [1/(x+h)^2  1/x^2]/h
ignoring the limit notation for now, we have
(x^2  (x+h)^2)/(hx^2(x+h)^2)
= (2hxh^2)/(hx^2(x+h)^2)
= 2(x+h)/(x^2(x+h)^2)
= 2/(x^2(x+h))
lim(h>0) is thus 2/x^3
at x=2, that's f'(2) = 1/4