1. ## Finding the derivative

Find the derivative of the following from first principles:

$
x^3 + 2x
$

2. Originally Posted by Mr Rayon
Find the derivative of the following from first principles:

$
x^3 + 2x
$
Do you know the appropriate formula to be using? Where do you get stuck in applying it? Please show what work you've done and where you get stuck.

(Aside: You will find many worked examples both at MHF and elsewhere).

3. Originally Posted by mr fantastic
Do you know the appropriate formula to be using? Where do you get stuck in applying it? Please show what work you've done and where you get stuck.
Very well...

$f(x) = x^3 - 2x$

$f(x + h) = (x + h)^3 + 2(x + h)$

$f(x + h) - f(x) = x^3 + h^3 - 2x - 2h$

$= x(x^2 - 2) + h(h^2 - 2)$

Substituting into, $\lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$, we get:

$\lim_{h\to 0} \frac{x(x^2 - 2) + h(h^2 - 2)}{h}$

$\lim_{h\to 0} x(x^2 - 2) + h^2 - 2, h\neq 0$

$= x ^3 - 2x - 2$

Originally Posted by mr fantastic
(Aside: You will find many worked examples both at MHF and elsewhere).
Oh really? Could you send me the links?

4. Originally Posted by Mr Rayon
Very well...

$f(x) = x^3 - 2x$

$f(x + h) = (x + h)^3 + 2(x + h)$

$f(x + h) - f(x) = x^3 + h^3 - 2x - 2h$

$= x(x^2 - 2) + h(h^2 - 2)$

Substituting into, $\lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$, we get:

$\lim_{h\to 0} \frac{x(x^2 - 2) + h(h^2 - 2)}{h}$

$\lim_{h\to 0} x(x^2 - 2) + h^2 - 2, h\neq 0$

$= x ^3 - 2x - 2$
That's not correct.

$f(x+h)-f(x)\neq x(x^2 - 2) + h(h^2 - 2)$!

$f(x+h)=(x+h)^3-2(x+h)=x^3+3x^2h+3xh^2+h^3-2x-2h$

Thus, $f(x+h)-f(x)=x^3+3x^2h+3xh^2+h^3-2x-2h-x^3+2x=3x^2h+3xh^2+h^3-2h$

So what is $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ equal to now?