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

Question

Ezekiel Alexander · Accepted Answer

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#

Emma Aldridge · Answer

undefined To find $ f'(x) $ for $ f(x) = 3x^2 + x $ using the limit definition of derivative, follow these steps:

1. Write down the limit definition of derivative:
$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

2. Substitute the given function $ f(x) = 3x^2 + x $ into the formula:
$$ f'(x) = \lim_{h \to 0} \frac{(3(x + h)^2 + (x + h)) - (3x^2 + x)}{h} $$

3. Expand and simplify the expression:
$$ f'(x) = \lim_{h \to 0} \frac{3(x^2 + 2xh + h^2) + x + h - (3x^2 + x)}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 + x + h - 3x^2 - x}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{6xh + 3h^2 + h}{h} $$
$$ f'(x) = \lim_{h \to 0} (6x + 3h + 1) $$

4. Now, plug in $ h = 0 $:
$$ f'(x) = 6x + 1 $$

Therefore, the derivative of $ f(x) = 3x^2 + x $ is $ f'(x) = 6x + 1 $.