Question

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

Answer 1

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

Explanation:

`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`