2x^2 - x - 8
i get to
4xdeltax + 2(deltax)^2 - deltax
the 2(deltax)^2 part is confusing me I get the answer 4x but it's supposed to be 4x - 1
thanks for any help
$\displaystyle \lim_{h\to 0}\frac{2(x+h)^{2}-(x+h)-8-(2x^{2}-x-8)}{h}$
Expanding the top and we get:
$\displaystyle \lim_{h\to 0}\frac{2x^{2}+4xh+2h^{2}-x-h-8-2x^{2}+x+8}{h}$
Note that many of the terms in the top cancel one another and we are left with:
$\displaystyle \lim_{h\to 0}\frac{4xh+2h^{2}-h}{h}$
$\displaystyle \lim_{h\to 0}4x+2h-1$
See what's left?. The derivative of $\displaystyle 2x^{2}-x-8$