f(x) = x^3 - x Find f'(x) using INCREMENT FORM of the DEFINITION of DERIVATIVE. Thanks.
f'(x)=[\lim_{h \rightarrow 0}\](f'(x+h)-f(x))/h
yeah okay see it through guys
i got 3x^2 + 3xh + h^2
my equation editor is no working. .
seems that way
shut up kiid i need a chain of equalities
Calyne, now put h=0 in ur final step
okay so it's 3x^2?
but isn't d/dx (x^3 - x) = 3x^2 - 1 ..?
I think you should calculate from the limit step again.
no, bro, i don't think so.. just fluttering show me
(x+h)^3-(x+h)-x^3+x x^3+3x^2h+3xh^2+h^3-x^3-x-h+x 3x^2+3xh+h^2-1(Since in the denominator there is a h) 3x^2-1
It should be \[f'(x) = \lim_{h \rightarrow 0} \frac {(x^3 + h)^3 - (x+h) - (x^3 - x)}{h}\]\[f'(x) = \lim_{h \rightarrow 0} \frac {-h + h^3 + 3h^2x + 3hx^2}{h}\]\[f'(x) = \lim_{h \rightarrow 0} = h^2 + 3hx + 3x^2 - 1 = 3x^2 - 1\]
Disregard the equal sign after the lim in the last... fail typing.
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