How do I find the derivative of (x^2+1)/(x-2) without using any derivative rules?

Using:
f'(x) = lim h->0 (f(x+h) - f(x))/h

Given:
f(x) = (x^2+1)/(x-2)

Apply definition to f(x):
f'(x) = lim h->0 ((((x+h)^2+1)/((x+h)-2)) - ((x^2+1)/(x-2))) / h

Multiply the terms and simplify:
f'(x) = lim h->0 ((((x^2 + 2xh +h^2)+1)/(x+h-2) - ((x^2+1)/(x-2))) / h
f'(x) = lim h->0 ((x^2 + 2xh +h^2 +1)/(x+h-2) - ((x^2+1)/(x-2))) / h
f'(x) = lim h->0 ((x^2 + 2xh +h^2 +1)/(h(x+h-2)) - ((x^2+1)/(h(x-2))))
f'(x) = lim h->0 (x^2 + 2xh +h^2 +1)/(hx+h^2-2h) - ((x^2+1)/(hx-2h))

Put the two subexpressions over common denominator:
f'(x) = lim h->0
((x^2 + 2xh +h^2 +1)*(hx-2h) -  ((x^2+1)*(hx+h^2-2h))) /
(hx-2h)*(hx+h^2-2h)

Multiply and simplify:
f'(x) = lim h->0
((x^3h - 2x^2h + 2x^2h^2 - 4xh^2 + xh^3 - 2h^3 + hx - 2h) -
( x^3h + x^2h^2 -2x^2h + (hx+h^2-2h ))) /
(x^2h^2 + xh^3 -2xh^2 -2xh^2 -2h^3 +4h^2)

Simplify by adding (thus canceling) opposite terms:
f'(x) = lim h->0
(x^3h - 2x^2h + 2x^2h^2 - 4xh^2 + xh^3 - 2h^3 + hx - 2h
-x^3h - x^2h^2 + 2x^2h - hx - h^2 + 2h ) /
(x^2h^2 + xh^3 -4xh^2 -2h^3 +4h^2)

Continue adding (thus canceling) opposite terms:
f'(x) = lim h->0
(2x^2h^2 - 4xh^2 + xh^3 - 2h^3 - x^2h^2 - h^2 ) /
(x^2h^2 + xh^3 -4xh^2 -2h^3 +4h^2)

Factor "h^2" from both numerator and denominator:
f'(x) = lim h->0 (2x^2 - 4x + xh - 2h - x^2 - 1 ) / (x^2 + xh -4x -2h +4)

Since there are no infinities left, take the limit:
f'(x) = (2x^2 - 4x + x*0 - 2*0 - x^2 - 1 ) / (x^2 + x*0 -4x -2*0 +4)
f'(x) = (2x^2 - 4x - x^2 - 1 ) / (x^2 -4x +4)
f'(x) = (x^2 - 4x - 1 ) / (x^2 -4x +4)

Factor denominator:
f'(x) = (x^2 - 4x - 1 ) / (x-2)^2



Now that we've determined the derivative using the definition,
we can at least use the quotient rule to *check* our work.
f'(x) = ((x-2)*2x - (x^2+1)*1) / (x-2)^2
f'(x) = (2x^2 -4x - x^2 - 1) / (x-2)^2
f'(x) = (x^2 -4x - 1) / (x-2)^2

Correct.



.