the derivative of f(x) = tcubed. I get the answer f'(x) = 3xcubed but I think the answer is supposed to be 3xsquared. (using the shortcut way). I'll put my workings below please wait and then please help!

\[\lim h \rightarrow 0 x - x ^{3} + 3x ^{3}h + 3xh ^{3} + h ^{3}/h\] \[\lim h \rightarrow 0 3x ^{3}h = 3xh ^{3} + h ^{3}/h\] \[\lim h \rightarrow 0 3x ^{3} + 3xh ^{2} + h ^{2}\] (make h 0) \[3x ^{3}\]

Oh why didn't that come up as equation... annoying...

x - xcubed + 3xcubedh + 3hcubedx + hcubed/h 3xcubedh + 3xhcubed + hcubed/h 3xcubed + 3xhsqured + hsquared make h 0 3xcubed

t^3 derives to 3t^2

this whole cubed, squared, english stuff is confusing; use the '^' for exponent

Hmmm yes that's what I thought, glad of the confirmation, but why can't I get that answer? Okay I'll do that.

\[\frac{(x+h)^3-(x^3)}{h}\]

x - x^3 + 3x^3h + 3h^3x + h^3/h 3x^3h + 3xh^3 + h^3/h 3x^3 + 3xh^2 + h^2 make h 0 3x^3

\(x^3+3x^2h+3xh^2+h^3-x^3\)//h \(3x^2h +3xh^2 +h^3\)//h \(3x^2 +3xh +h^2\) ; when h=0 \(3x^2\)

I got the binomial theory wrong! Yes thankyou very much!

:) i saw that

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