What do you want, the derivative?
if so
f(x) = 1/x
f(x+h) = 1/(x+h)
f(x+h)-f(x) = 1/(x+h) - 1/x
= [ x -(x+h)] /[x(x+h)]
= -h/ (x^2 + x h)
[f(x+h)-f(x)]/h = -1/(x^2+xh)
limit as h-->0
= -1/x^2
which is of course the derivative
1/X by first principles
4 answers
Where did -h come from
h is a small increase in x which is allowed to approach zero to get the slope of f(x) at x, the definition of the derivative
Definition of deriviative of f(x) is:
df/dx = {f(x+h) - f(x) ]/h as h-->0
Definition of deriviative of f(x) is:
df/dx = {f(x+h) - f(x) ]/h as h-->0
Oh, if you mean the algebra look at the numerator
[ x -(x+h)] = x - x - h
= -h
[ x -(x+h)] = x - x - h
= -h