| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | First principles: x² terms |
| Difficulty | Moderate -0.8 This is a straightforward application of the first principles definition to differentiate a simple power term (x²). While it requires careful algebraic manipulation, it's a standard textbook exercise with a well-rehearsed method and no conceptual challenges beyond recalling the definition and basic algebra. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{2(x+h)^2 - 2x^2}{h} = \ldots\) | M1 | Writes gradient of chord and attempts to expand bracket; condone poor squaring e.g. \((x+h)^2=x^2+h^2\) |
| \(\dfrac{2(x+h)^2-2x^2}{h} = \dfrac{4xh+2h^2}{h}\) | A1 | Reaches correct fraction with \(x^2\) terms cancelled |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h\to 0}\dfrac{4xh+2h^2}{h} = \lim_{h\to 0}(4x+2h) = 4x\) | A1* | Must apply limiting argument and deduce \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=4x\); must reach \(4x+2h\) before limit; \(h\to0\) must appear somewhere; no errors seen |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{2(x+h)^2 - 2x^2}{h} = \ldots$ | M1 | Writes gradient of chord and attempts to expand bracket; condone poor squaring e.g. $(x+h)^2=x^2+h^2$ |
| $\dfrac{2(x+h)^2-2x^2}{h} = \dfrac{4xh+2h^2}{h}$ | A1 | Reaches correct fraction with $x^2$ terms cancelled |
| $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h\to 0}\dfrac{4xh+2h^2}{h} = \lim_{h\to 0}(4x+2h) = 4x$ | A1* | Must apply limiting argument and deduce $\dfrac{\mathrm{d}y}{\mathrm{d}x}=4x$; must reach $4x+2h$ before limit; $h\to0$ must appear somewhere; no errors seen |\begin{enumerate}
\item Given that
\end{enumerate}
$$y = 2 x ^ { 2 }$$
use differentiation from first principles to show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x$$
\hfill \mbox{\textit{Edexcel Paper 2 2022 Q4 [3]}}