Ho do I use the limit definition of derivative to find f'(x) for f(x)=3x^2+x ?

1 Answer
Aug 30, 2014

By Power Rule, we know that we are supposed to get f'(x)=6x+1.

Let us find it using the definition
f'(x)=lim_{h to 0}{f(x+h)-f(x)}/{h}

Let us first find the difference quotient
{f(x+h)-f(x)}/{h}={3(x+h)^2+(x+h)-[3x^2+x]}/{h}
By simplifying the numerator,
={3(x^2+2xh+h^2)+x+h-3x-x}/{h}
={3x^2+6xh+3h^2+h-3x^2}/{h}
={6xh+3h^2+h}/{h}
By factoring out h from the numerator,
={h(6x+3h+1)}/h
By cancelling out h's,
=6x+3h+1

Hence,
f'(x)=lim_{h to 0}(6x+3h+1)=6x+3(0)+1=6x+1