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

Aug 30, 2014

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

Let us find it using the definition
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

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

Hence,
$f ' \left(x\right) = {\lim}_{h \to 0} \left(6 x + 3 h + 1\right) = 6 x + 3 \left(0\right) + 1 = 6 x + 1$