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Old September 7th, 2008, 01:59 PM
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Default need help with algebra finding derivative

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
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Old September 7th, 2008, 02:09 PM
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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
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Old September 7th, 2008, 02:11 PM
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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
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Old September 7th, 2008, 02:15 PM
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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.
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Old September 7th, 2008, 02:16 PM
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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 ?
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Old September 7th, 2008, 02:17 PM
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doesnt -3(x+Ax) = -3x + -3Ax ? and then the other -3x so -6x
what other -3x? the only other 3x is +3x
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Old September 7th, 2008, 02:33 PM
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yeah i see what i did

but i still dont have the answer do i? look at my post above yours
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Old September 7th, 2008, 03:19 PM
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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 ...?
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