f(x) = x^3 - x; using increment form of the definition of derivative, find f'(x).

Question

anonymous · Answer

as in 
$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$??

anonymous · Answer

i am not familiar with the term "increment form"

anonymous · Answer

yes that one pretty sure

anonymous · Answer

PLEASE and THANK YOU

anonymous · Answer

ok lets to it

anonymous · Answer

$$f(x)=x^3-x$$
$$f(x+h)=(x+h)^3-(x+h)$$ then algebra 
$$f(x+h)=(x+h)^3-(x+h)=x^3+3x^2h+3xh^2+h^2-x-x$$

anonymous · Answer

damn typo 
$$f(x+h)=(x+h)^3-(x+h)=x^3+3x^2h+3xh^2+h^2-x-h$$

anonymous · Answer

now 
$$f(x+h)-f(x)=x^3+3x^2h+3xh^2+h^2-x-h-(x^3-x)$$
$$=3x^2h+3xh^2+h^3-h$$

anonymous · Answer

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

anonymous · Answer

hope all the steps are there
i see the binary dust is coming, so i will go watch judge judy