Question

(2(x+h)^3-2x^3)/h

Answer 1

got to do a bunch of algebra here to cube
\[(x+h)^3=x^3+3x^3h+3xh^2+h^3\]

Answer 2

so you get
\[\frac{2(x^3+3x^2h+3xh^2+h^3)-2x^3}{h}=\frac{6x^2h+6xh^2+2h^3}{h}\]
\[=\frac{h(6x^2+6xh+2h^2)}{h}=6x^2+6xh+2h^2\]

Answer 3

ya i got that far but breaking it down from there is messing me up I'm suppose to end up with 6x^2 for the answer

Answer 4

? Probably because you are finding derivatives, and as h-> 0, 6xh and 2h^2 become 0, leaving you with 6x^2 ?

Answer 5

if you replace h by zero at the end you are only left with
\[6x^2\]

Answer 6

got it thank you for the help