f(x) = x^3 - x; using increment form of the definition of derivative, find f'(x).
as in \[f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\]??
i am not familiar with the term "increment form"
yes that one pretty sure
PLEASE and THANK YOU
ok lets to it
\[f(x)=x^3-x\] \[f(x+h)=(x+h)^3-(x+h)\] then algebra \[f(x+h)=(x+h)^3-(x+h)=x^3+3x^2h+3xh^2+h^2-x-x\]
damn typo \[f(x+h)=(x+h)^3-(x+h)=x^3+3x^2h+3xh^2+h^2-x-h\]
now \[f(x+h)-f(x)=x^3+3x^2h+3xh^2+h^2-x-h-(x^3-x)\] \[=3x^2h+3xh^2+h^3-h\]
and finally \[\frac{f(x+h)-f(x)}{h}=\frac{=3x^2h+3xh^2+h^3-h}{h}=\frac{h(3x^2+3xh+h^2-1)}{h}\] \[=3x^2+3xh+h^2-1\] now take the limit as h goes to zero (be replacing h by zero) and you get \[3x^2-1\] just like it should be
hope all the steps are there i see the binary dust is coming, so i will go watch judge judy
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