How do you use the definition of a derivative to find the derivative of #f(x)=2x^2+5x#?

1 Answer
Apr 27, 2017

Answer:

# f'(x ) = 4x+5#

Explanation:

By definition of the derivative:

# f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h #

So with # f(x) = 2x^2+5x # we have;

# f'(x ) = lim_(h rarr 0) ( {2(x+h)^2+5(x+h)} -{2x^2+5x} ) / h #
# " " = lim_(h rarr 0) ( {2(x^2+2hx+h^2)+5x+5h} -{2x^2+5x} ) / h #
# " " = lim_(h rarr 0) ( {2x^2+4hx+2h^2+5x+5h} -{2x^2+5x} ) / h #
# " " = lim_(h rarr 0) ( 2x^2+4hx+2h^2+5x+5h - 2x^2-5x ) / h #
# " " = lim_(h rarr 0) ( 4hx+2h^2+5h ) / h #
# " " = lim_(h rarr 0) ( 4x+2h+5) #
# " " = 4x+5#