How do you use the limit definition to find the derivative of f(x)=2x^2+1f(x)=2x2+1?

1 Answer
Steve M
Nov 2, 2016

f'(x) = 4x

Explanation:

By definition, f'(x)= lim_(h->0)(f(x+h)-f(x))/h

so for f(x) = 2x^2 +1 we have:
f'(x) = lim_(h->0)( { (2(x+h)^2 +1) - (2x^2 +1)} )/h
f'(x) = lim_(h->0)( (2(x^2+2hx+h^2) +1 ) - 2x^2+1)/h
f'(x) = lim_(h->0)( 2x^2+4hx+2h^2 +1 - 2x^2-1 )/h
f'(x) = lim_(h->0)( 4hx+2h^2 )/h
f'(x) = lim_(h->0)( 4x+2h)
f'(x) = 4x

Related questions
See all questions in Limit Definition of Derivative
Impact of this question
14121 views around the world
You can reuse this answer
Creative Commons License