Question

Find the derivative of f(x)=x^3-2 using f'(x)= lim h->0 f(x+h)-f(x)/h.

Answer 1

\[f(x)=x^3-2\\
f'(x)=\lim_{h\to0}\frac{\bigg((x+h)^3-2\bigg)-\bigg(x^3-2\bigg)}{h}\]
Expand the numerator. You'll find that all the terms not containing \(h\) will disappear, leaving you with a few terms that do contain it.

Answer 2

So take h out? and go from there?

Answer 3

Not until you've expanded the numerator:
\[(x+h)^3-2-(x^3-2)=x^3+3x^2h+3xh^2+h^3-2-x^3+2\]which simplifies to \[3x^2h+3xh^2+h^3\]

Answer 4

where do you get the 3 from?

Answer 5

It's a byproduct of the expansion:
\[(x+h)^3=x^3+3x^2h+3xh^2+h^3\]

Answer 6

okay so i take the h out and get \[3x ^{2}+3xh+h ^{2}\] where do i go from there?

Answer 7

Yep, next thing is to take the limit:
\[\lim_{h\to0}(3x^2+3xh+h^2)=\cdots\]

Answer 8

so \[3x ^{2}+3x\]?

Answer 9

Close, the second term also disappears, since it contains a factor of \(h\). You're left with just \(3x^2\).

Answer 10

so it goes away even though there is an x there?

Answer 11

Yep, the limit only takes \(h\) into consideration (since it is the limit as \(h\to0\), not \(x\)).

Answer 12

You can treat \(x\) as any old constant in this context.

Answer 13

Oh alright cool thank you!

Answer 14

You're welcome