Thread: need help with algebra finding derivative
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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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