I'm working on this question:
if f(x) = x^3, then the derivative is 3x^2
I have to prove this using the definition of the derivative. I got as far as this:
I'm unsure of how to simplify it in order to find the limit as h -> 0.
I'm working on this question:
if f(x) = x^3, then the derivative is 3x^2
I have to prove this using the definition of the derivative. I got as far as this:
I'm unsure of how to simplify it in order to find the limit as h -> 0.
lim h --> 0 of x^3 = [(x+h)^3 - x^3]/h
= [(x+h)(x+h)(x+h) - x^3]/h
= [x^3 + x^2 h + 2 x^2 h + 2 x h^2 + x h^2 + h^3 - x^3]/h
cancel the x^3 and the h's with the one in the bottom
= x^2 + 2 x^2 + 2 x h + x h + h^2
plug in 0 for h
= x^2 + 2 x^2
combine like terms
= 3 x^2
I hope that helps