| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2014 |
| Session | June |
| Marks | 4 |
| Topic | Differentiation from First Principles |
| Type | First principles: x³ terms |
| Difficulty | Moderate -0.8 This is a standard textbook exercise requiring recall of the first principles formula and algebraic manipulation of (x+h)³ using the binomial expansion. While it requires careful algebra, it's a routine application with no problem-solving insight needed, making it easier than average for A-level. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x |
$(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ seen anywhere
B1 Or unsimplified equiv. Could expand $(x-h)^3$ instead
If $f(x) = x^3$, $f(x+h) = (x+h)^3$
M1 Just recognise that $f(x+h) = (x+h)^3$, or $f(x-h) = (x-h)^3$
$\dfrac{f(x+h) - f(x)}{h} = \dfrac{(x+h)^3 - x^3}{h}$
$= 3x^2 + 3xh + h^2$
M1 Attempt correct process, including division by $h$
then $f'(x) = \lim_{h \to 0}(3x^2 + 3xh + h^2) = 3x^2$
A1 [4] Allow $h = 0$ for $h \to 0$. Allow $f'(x) \to 3x^2$. Need to see $f'(x)$ or $\dfrac{dy}{dx}$ within proof3 Given that $\mathrm { f } ( x ) = x ^ { 3 }$, use differentiation from first principles to prove that $\mathrm { f } ^ { \prime } ( x ) = 3 x ^ { 2 }$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2014 Q3 [4]}}