| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| 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 standard textbook exercise requiring recall of the first principles formula and application of a basic algebraic identity (a²-b²). While it requires careful algebraic manipulation, it's a routine question with no problem-solving element—students practice this exact derivation repeatedly when learning differentiation from first principles. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{(x+h)^2 - x^2}{h} = \frac{x^2 + 2xh + h^2 - x^2}{h}\) | M1 | Begins by writing gradient of chord and attempts to expand the squared bracket; the \(-x^2\) term must be present |
| \(= \frac{2xh + h^2}{h}\) | A1 | Reaches correct fraction with \(x^2\) terms cancelled, no algebraic errors |
| \(\frac{dy}{dx} = \lim_{h \to 0} \frac{2xh+h^2}{h} = \lim_{h \to 0}(2x+h) = 2x\) | A1* | Completes with limiting argument, deduces \(\frac{dy}{dx} = 2x\); must have \(= 2x\) not just \(\lim_{h\to 0} 2x\); \(\frac{dy}{dx}=\) or equivalent must appear |
| Total | (3) |
# Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{(x+h)^2 - x^2}{h} = \frac{x^2 + 2xh + h^2 - x^2}{h}$ | M1 | Begins by writing gradient of chord and attempts to expand the squared bracket; the $-x^2$ term must be present |
| $= \frac{2xh + h^2}{h}$ | A1 | Reaches correct fraction with $x^2$ terms cancelled, no algebraic errors |
| $\frac{dy}{dx} = \lim_{h \to 0} \frac{2xh+h^2}{h} = \lim_{h \to 0}(2x+h) = 2x$ | A1* | Completes with limiting argument, deduces $\frac{dy}{dx} = 2x$; must have $= 2x$ not just $\lim_{h\to 0} 2x$; $\frac{dy}{dx}=$ or equivalent must appear |
| **Total** | **(3)** | |
---\begin{enumerate}
\item Given that $y = x ^ { 2 }$, use differentiation from first principles to show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 x$
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q4 [3]}}