Expand ${(x + h)^3}$.

Hence find the derivative of $f(x) = {x^3}$ from first principles.

${(x + h)^3} = {x^3} + 3{x^2}h + 3x{h^2} + {h^3}$     (M1)A1

[2 marks]

$f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{{(x + h)}^3} - {x^3}}}{h}$     (M1)

$= \mathop {\lim }\limits_{h \to 0} \frac{{{x^3} + 3{x^2}h + 3x{h^2} + {h^3} - {x^3}}}{h}$

$= \mathop {\lim }\limits_{h \to 0} (3{x^2} + 3xh + {h^2})$     A1

$= 3{x^2}$     A1

Note:     Do not award final A1 on FT if $= 3{x^2}$ is not obtained

Note:     Final A1 can only be obtained if previous A1 is given

[3 marks]

Total [5 marks]

Well done although some did not use the binomial expansion.

Fine by those who knew what first principles meant, not by the others.