Using the limit definition, how do you find the derivative of #F(x)=x^3−7x+5#?

Question

Evelyn Agard · Accepted Answer

#f'(x) = 3x^2 -7#

#f(x)=x^3−7x+5#

by definition

#f'(x) = lim_{h \to 0} (f(x+h) - f(x))/(h)#

#f'(x) = lim_{h \to 0} ( {(x+h)^3−7(x+h)+5}- {x^3−7x+5})/(h)#

when expanding out that binomial, Pascal's Triangle is handy to know.

#f'(x) = lim_{h \to 0} ( {(x^3+3x^2h +3xh^2 +h^3)−7(x+h)+5}- {x^3-7x+5})/(h)#

#f'(x) = lim_{h \to 0} ( x^3+3x^2h +3xh^2 +h^3−7x-7h+5- x^3+7x-5)/(h)#

#f'(x) = lim_{h \to 0} ( 3x^2h +3xh^2 +h^3-7h)/(h)#

because we are looking at the limit #h \to 0# such that #h \ne 0#, we can do normal algebra and cancel the h's where appropriate

#f'(x) = lim_{h \to 0} 3x^2 +3xh +h^2-7#

#f'(x) = lim_{h \to 0} 3x^2 -7 + \mathcal{O}(h)#

#f'(x) = 3x^2 -7#

William Abdalla · Answer

undefined To find the derivative of $ F(x) = x^3 - 7x + 5 $ using the limit definition, we use the formula for the derivative:

$$ f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h} $$

First, we substitute $ f(x) = x^3 - 7x + 5 $ into the formula:

$$ f'(x) = \lim_{{h \to 0}} \frac{{(x + h)^3 - 7(x + h) + 5 - (x^3 - 7x + 5)}}{h} $$

Next, we expand and simplify the numerator:

$$ f'(x) = \lim_{{h \to 0}} \frac{{x^3 + 3x^2h + 3xh^2 + h^3 - 7x - 7h + 5 - x^3 + 7x - 5}}{h} $$

Combine like terms and cancel out the common terms:

$$ f'(x) = \lim_{{h \to 0}} \frac{{3x^2h + 3xh^2 + h^3 - 7h}}{h} $$

Now, factor out an $ h $ from the numerator:

$$ f'(x) = \lim_{{h \to 0}} \frac{{h(3x^2 + 3xh + h^2 - 7)}}{h} $$

Cancel out the $ h $ in the numerator and denominator:

$$ f'(x) = \lim_{{h \to 0}} (3x^2 + 3xh + h^2 - 7) $$

Now, substitute $ h = 0 $ into the expression:

$$ f'(x) = 3x^2 - 7 $$

So, the derivative of $ F(x) = x^3 - 7x + 5 $ using the limit definition is $ f'(x) = 3x^2 - 7 $.