need help with algebra finding derivative

stones44 · #1 $Old$ 09-07-2008, 01:59 PM

Show that the slope of y = x^3 - 3x + 3 at any x is 3x^2 - 3

A=delta (dont know how to do delta)

im doing it using this method
( (x+Ax)^3 - 3(x+Ax) + 3 ) - ( x^3 - 3x + 3 ) / ( (x+Ax) - x )

i simplify and get Ax^2 + 3xAx + 3x^2 - 6x

Jhevon · #2 $Old$ 09-07-2008, 02:09 PM

Quote:

Originally Posted by stones44 $View Post$

Show that the slope of y = x^3 - 3x + 3 at any x is 3x^2 - 3

A=delta (dont know how to do delta)

im doing it using this method
( (x+Ax)^3 - 3(x+Ax) + 3 ) - ( x^3 - 3x + 3 ) / ( (x+Ax) - x )

i simplify and get Ax^2 + 3xAx + 3x^2 - 6x

where did 6x come from?

you simplified something wrong. expanding everything, you should get:

$\lim_{\Delta x \to 0} \frac {x^3 + 3x^2(\Delta x) + 3x(\Delta x)^2 + (\Delta x)^3 - 3x - 3 \Delta x + 3 - x^3 + 3x - 3}{\Delta x}$

now cancel what is to be canceled, simplify as much as possible, and take the limit

stones44 · #3 $Old$ 09-07-2008, 02:11 PM

Quote:

Originally Posted by Jhevon $View Post$

where did 6x come from?

you simplified something wrong. expanding everything, you should get:

$\lim_{\Delta x \to 0} \frac {x^3 + 3x^2(\Delta x) + 3x(\Delta x)^2 + (\Delta x)^3 - 3x - 3 \Delta x + 3 - x^3 + 3x - 3}{\Delta x}$

now cancel what is to be canceled, simplify as much as possible, and take the limit

doesnt -3(x+Ax) = -3x + -3Ax ? and then the other -3x so -6x

skeeter · #4 $Old$ 09-07-2008, 02:15 PM

let $h = \Delta x$

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

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

$\lim_{h \to 0} \frac{(x+h)^3 - 3(x+h) + 3 - (x^3 - 3x + 3)}{h}$

$\lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - 3x - 3h + 3 - (x^3 - 3x + 3)}{h}$

now, clean it up and finish finding the limit which will give you the derivative of f.

stones44 · #5 $Old$ 09-07-2008, 02:16 PM

i figured what i did wrong

ok so i cleaned it up i have

3x^2 + 3xAx + Ax^2 - 3

shouldnt i end up with (3x^2 - 3) +Ax ?

Jhevon · #6 $Old$ 09-07-2008, 02:17 PM

Quote:

Originally Posted by stones44 $View Post$

doesnt -3(x+Ax) = -3x + -3Ax ? and then the other -3x so -6x

what other -3x? the only other 3x is +3x

stones44 · #7 $Old$ 09-07-2008, 02:33 PM

yeah i see what i did

but i still dont have the answer do i? look at my post above yours

Jhevon · #8 $Old$ 09-07-2008, 03:19 PM

Quote:

Originally Posted by stones44 $View Post$

yeah i see what i did

but i still dont have the answer do i? look at my post above yours

nope, you should end up with $3x^2 + 3x (\Delta x) + (\Delta x)^2 - 3$

and if we let $\Delta x \to 0$ ...?