Question

Prove that the derivative of \[x^3\]is \[3x^2\]

Answer 1

Is this like your "prove 2+2=4" question, or more of a "limit definition of derivative" question?

Answer 2

limit definition

Answer 3

?

Answer 4

\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}=\lim_{h \rightarrow 0}\frac{(x+h)^{3}-x^{3}}{h}\]Numerator:\[(x+h)^{3}-x^{3} = x^{3}+3x^{2}h+3xh^{2}+h^{3} - x^{3} = 3x^{2}h+3xh^{2}+h^{3}\]\[\lim_{h \rightarrow 0}\frac{3x^{2}h+3xh^{2}+h^{3}}{h} = \lim_{h \rightarrow 0}(3x^{2}+3xh+h^{2})=3x^{2}\]

Answer 5

is it still true even if x is very large? how would you ignore the 3xh term

Answer 6

show that the error\[3xh + h^2\]can be made arbitrarily small

Answer 7

mm i would go :
3x^2 / 3xh as x->infinity
3x / 3h -> infinity
so.. it is much much bigger than 3xh

Answer 8

You ignore 3xh because h goes to 0.

As for the error, I think that's something like show that for an arbitrarily small epsilon you can find h such that\[\left| 3xh+h^{2} \right|<\epsilon\]So\[-\epsilon < 3xh+h^{2}<\epsilon\]So solve\[h^{2}+3xh-\epsilon=0\]for h:\[h=\frac{-3x\pm \sqrt{9x^{2}+4\epsilon}}{2}\]Something, something, something. Choose h smaller than some function of your huge x and tiny epsilon. That didn't quite work.

Answer 9

that works; as long as h can be made arbitrarily small, it works.

Answer 10

But I'm picking some h that's directly proportional to x. Can't be right. But that's sorta the process. It's been over 20 years. I got the limit part right. What do you expect from me?!?! ;)

Answer 11

remember x is fixed in this process, so we have no trouble taking the limit.