| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | First principles: polynomial (find gradient) |
| Difficulty | Moderate -0.5 This is a straightforward application of the first principles formula to a simple polynomial (cubic). While it requires careful algebraic manipulation of the difference quotient and factoring (x+h)³, it's a standard textbook exercise with no conceptual surprises. Slightly easier than average because the polynomial is simple and the method is purely procedural. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of chord = \(\frac{(2(x+h)^3+5)-(2x^3+5)}{x+h-x}\) | B1 | \(2(x+h)^3+5\), seen or implied |
| Attempts to write gradient of chord in terms of \(x\) and \(h\) | M1 | Begins the proof |
| \((x+h)^3 = x^3+3x^2h+3xh^2+h^3\) | B1 | By expanding or correct binomial expansion |
| \(= \frac{2x^3+6x^2h+6xh^2+2h^3+5-2x^3-5}{h}\) | ||
| \(= 6x^2+6xh+2h^2\) | A1 | Correct gradient of chord |
| \(\frac{dy}{dx} = \lim_{h\to 0}(6x^2+6xh+2h^2)=6x^2\) and so at \(P\), \(\frac{dy}{dx}=6(1)^2=6\) | A1 | Applies limiting argument; deduces \(\frac{dy}{dx}=6\) at \(P\) |
| Total: (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(Q\) have \(x\) coordinate \(1+h\), so \(y_Q=2(1+h)^3+5\) | B1 | Writes \(y\) coordinate of point close to \(P\) |
| Gradient \(PQ = \frac{2(1+h)^3+5-7}{1+h-1}\) | M1 | Attempts gradient of chord \(PQ\) in terms of \(h\) |
| \((1+h)^3=1+3h+3h^2+h^3\) | B1 | By expanding or correct binomial expansion |
| \(= \frac{6h+6h^2+2h^3}{h} = 6+6h+2h^2\) | A1 | Correct gradient of chord |
| \(\frac{dy}{dx}=\lim_{h\to 0}(6+6h+2h^2)=6\) | A1 | Applies limiting argument to deduce \(\frac{dy}{dx}=6\) at \(P\) |
| Total: (5) |
## Question 9:
**Main Method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of chord = $\frac{(2(x+h)^3+5)-(2x^3+5)}{x+h-x}$ | B1 | $2(x+h)^3+5$, seen or implied |
| Attempts to write gradient of chord in terms of $x$ and $h$ | M1 | Begins the proof |
| $(x+h)^3 = x^3+3x^2h+3xh^2+h^3$ | B1 | By expanding or correct binomial expansion |
| $= \frac{2x^3+6x^2h+6xh^2+2h^3+5-2x^3-5}{h}$ | | |
| $= 6x^2+6xh+2h^2$ | A1 | Correct gradient of chord |
| $\frac{dy}{dx} = \lim_{h\to 0}(6x^2+6xh+2h^2)=6x^2$ and so at $P$, $\frac{dy}{dx}=6(1)^2=6$ | A1 | Applies limiting argument; deduces $\frac{dy}{dx}=6$ at $P$ |
| **Total: (5)** | | |
**Alt 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $Q$ have $x$ coordinate $1+h$, so $y_Q=2(1+h)^3+5$ | B1 | Writes $y$ coordinate of point close to $P$ |
| Gradient $PQ = \frac{2(1+h)^3+5-7}{1+h-1}$ | M1 | Attempts gradient of chord $PQ$ in terms of $h$ |
| $(1+h)^3=1+3h+3h^2+h^3$ | B1 | By expanding or correct binomial expansion |
| $= \frac{6h+6h^2+2h^3}{h} = 6+6h+2h^2$ | A1 | Correct gradient of chord |
| $\frac{dy}{dx}=\lim_{h\to 0}(6+6h+2h^2)=6$ | A1 | Applies limiting argument to deduce $\frac{dy}{dx}=6$ at $P$ |
| **Total: (5)** | | |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = 2 x ^ { 3 } + 5$$
The curve $C$ passes through the point $P ( 1,7 )$.\\
Use differentiation from first principles to find the value of the gradient of the tangent to $C$ at $P$.
\hfill \mbox{\textit{Edexcel Paper 1 Q9 [5]}}