Question 880471
Previously Posted
derivative of f(x)= 2x^2+3x-5 
f'(x) = lim {{{(f(x+h) - f(x))/h}}}  as h approaches 0
Plug and Play
{{{( 2(x+h)^2 + 3(x+h) - 5 - (2h^2 + 3h - 5))/h = (2x^2 + 4xh + 2h^2 +3x + 3h - 5 -2x^2 - 3x + 5)/h = (4xh + 2h^2 -3x)/h = (4xh + 2h^2 + 3h)/h = 4x + 3 + 2h }}}
{{{lim( h->0, (4x + 3 + 2h) ) = 4x+3 }}}