#f(x)=x^3−7x+5#
by definition
#f'(x) = lim_{h \to 0} (f(x+h) - f(x))/(h)#
#f'(x) = lim_{h \to 0} ( {(x+h)^3−7(x+h)+5}- {x^3−7x+5})/(h)#
when expanding out that binomial, Pascal's Triangle is handy to know.
#f'(x) = lim_{h \to 0} ( {(x^3+3x^2h +3xh^2 +h^3)−7(x+h)+5}- {x^3-7x+5})/(h)#
#f'(x) = lim_{h \to 0} ( x^3+3x^2h +3xh^2 +h^3−7x-7h+5- x^3+7x-5)/(h)#
#f'(x) = lim_{h \to 0} ( 3x^2h +3xh^2 +h^3-7h)/(h)#
because we are looking at the limit #h \to 0# such that #h \ne 0#, we can do normal algebra and cancel the h's where appropriate
#f'(x) = lim_{h \to 0} 3x^2 +3xh +h^2-7#
#f'(x) = lim_{h \to 0} 3x^2 -7 + \mathcal{O}(h)#
#f'(x) = 3x^2 -7#
