How do I use the limit definition of derivative to find #f'(x)# for #f(x)=5x-9x^2# ?

Question

Emily Ammerman · Accepted Answer

Let us find #f'(x)# by using the limit definition. Start with #f(x)#. #f(x)=5x-9x^2# Let us find #f(x+h)#. #f(x+h)=5(x+h)-9(x+h)^2#
#=5x+5h-9(x^2+2xh+h^2)#
#=5x+5h-9x^2-18xh-9h^2# Let us find the difference quotient. #{f(x+h)-f(x)}/h# by plugging in the expression we found above, #={5x+5h-9x^2-18xh-9h^2-(5x-9x^2)}/h# by cancelling out #5x#'s and #-9x^2#'s, #={5h-18xh-9h^2}/h# by factoring #h# out in the numerator, #={h(5-18x-9h)}/h# by cancelling out #h#'s, #=5-18x-9h# Now, we can find #f'(x)#. #f'(x)=lim_{h to 0}(5-18x-9h)=5-18x-9(0)=5-18x#

Emily Ammerman · Answer

undefined To find $ f'(x) $ using the limit definition of the derivative for $ f(x) = 5x - 9x^2 $, follow these steps:

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

2. Substitute the function $ f(x) = 5x - 9x^2 $ into the limit definition:
$$ f'(x) = \lim_{h \to 0} \frac{(5(x + h) - 9(x + h)^2) - (5x - 9x^2)}{h} $$

3. Expand and simplify the expression:
$$ f'(x) = \lim_{h \to 0} \frac{5x + 5h - 9(x^2 + 2xh + h^2) - 5x + 9x^2}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{5h - 9x^2 - 18xh - 9h^2}{h} $$
$$ f'(x) = \lim_{h \to 0} (5 - 18x - 9h) $$

4. Evaluate the limit as $ h $ approaches 0:
$$ f'(x) = 5 - 18x $$

Therefore, the derivative $ f'(x) $ of the function $ f(x) = 5x - 9x^2 $ is $ 5 - 18x $.