f(x) = x^3 - x
Find f'(x) using INCREMENT FORM of the DEFINITION of DERIVATIVE. Thanks.

Question

f(x) = x^3 - x 
Find f'(x) using INCREMENT FORM of the DEFINITION of DERIVATIVE. Thanks.

anonymous · Answer

f'(x)=[\lim_{h \rightarrow 0}\](f'(x+h)-f(x))/h

anonymous · Answer

yeah okay see it through guys

anonymous · Answer

i got 3x^2 + 3xh + h^2

anonymous · Answer

my equation editor is no working. .

anonymous · Answer

seems that way

anonymous · Answer

shut up kiid i need a chain of equalities

anonymous · Answer

Calyne, now put h=0 in ur final step

anonymous · Answer

okay so it's 3x^2?

anonymous · Answer

but isn't d/dx (x^3 - x) = 3x^2 - 1 ..?

anonymous · Answer

I think you should calculate from the limit step again.

anonymous · Answer

no, bro, i don't think so.. just fluttering show me

anonymous · Answer

(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

rogue · Answer

It should be $$f'(x) = \lim_{h \rightarrow 0} \frac {(x^3 + h)^3 - (x+h) - (x^3 - x)}{h}$$$$f'(x) = \lim_{h \rightarrow 0} \frac {-h + h^3 + 3h^2x + 3hx^2}{h}$$$$f'(x) = \lim_{h \rightarrow 0} = h^2 + 3hx + 3x^2 - 1 = 3x^2 - 1$$

rogue · Answer

Disregard the equal sign after the lim in the last... fail typing.