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)
= (-2hx-h^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