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

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##