# 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##.

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

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