Find the derivative of the following from first principles:
$\displaystyle
x^3 + 2x
$
Very well...
$\displaystyle f(x) = x^3 - 2x$
$\displaystyle f(x + h) = (x + h)^3 + 2(x + h)$
$\displaystyle f(x + h) - f(x) = x^3 + h^3 - 2x - 2h$
$\displaystyle = x(x^2 - 2) + h(h^2 - 2)$
Substituting into, $\displaystyle \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$, we get:
$\displaystyle \lim_{h\to 0} \frac{x(x^2 - 2) + h(h^2 - 2)}{h}$
$\displaystyle \lim_{h\to 0} x(x^2 - 2) + h^2 - 2, h\neq 0$
$\displaystyle = x ^3 - 2x - 2$
Oh really? Could you send me the links?
That's not correct.
$\displaystyle f(x+h)-f(x)\neq x(x^2 - 2) + h(h^2 - 2)$!
$\displaystyle f(x+h)=(x+h)^3-2(x+h)=x^3+3x^2h+3xh^2+h^3-2x-2h$
Thus, $\displaystyle f(x+h)-f(x)=x^3+3x^2h+3xh^2+h^3-2x-2h-x^3+2x=3x^2h+3xh^2+h^3-2h$
So what is $\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ equal to now?