| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | First principles: trigonometric functions |
| Difficulty | Standard +0.8 This is a standard first principles proof requiring the compound angle formula and two standard limits, but it demands careful algebraic manipulation and understanding of the limit definition. While it's a classic A-level question, proving derivatives from first principles is conceptually more demanding than routine differentiation, placing it moderately above average difficulty. |
| Spec | 1.07h Differentiation from first principles: for sin(x) and cos(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{\sin(x+h) - \sin x}{h}\) | B1 | Gives correct fraction. May be implied by \(\frac{\sin x \cos h + \cos x \sin h - \sin x}{h}\). Condone invisible brackets. |
| \(\dfrac{\sin x \cos h + \cos x \sin h - \sin x}{h}\) | M1 | Uses compound angle formula for \(\sin(x \pm h)\) to give \(\sin x \cos h \pm \cos x \sin h\) |
| \(\dfrac{\sin x \cos h + \cos x \sin h - \sin x}{h}\) | A1 | Achieves this expression or equivalent. Allow invisible brackets to be recovered. |
| \((\text{As } h \to 0),\ \sin x\!\left(\dfrac{\cos h - 1}{h}\right) + \cos x\!\left(\dfrac{\sin h}{h}\right) \to 0 \times \sin x + 1 \times \cos x\) | dM1 | Dependent on both B and M marks. Complete attempt to apply limits: must isolate \(\left(\frac{\cos h -1}{h}\right)\) and replace with 0, and isolate \(\left(\frac{\sin h}{h}\right)\) and replace with 1. |
| so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \cos x\) * | A1* | Printed answer — must be fully and correctly justified |
## Question 12:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{\sin(x+h) - \sin x}{h}$ | B1 | Gives correct fraction. May be implied by $\frac{\sin x \cos h + \cos x \sin h - \sin x}{h}$. Condone invisible brackets. |
| $\dfrac{\sin x \cos h + \cos x \sin h - \sin x}{h}$ | M1 | Uses compound angle formula for $\sin(x \pm h)$ to give $\sin x \cos h \pm \cos x \sin h$ |
| $\dfrac{\sin x \cos h + \cos x \sin h - \sin x}{h}$ | A1 | Achieves this expression or equivalent. Allow invisible brackets to be recovered. |
| $(\text{As } h \to 0),\ \sin x\!\left(\dfrac{\cos h - 1}{h}\right) + \cos x\!\left(\dfrac{\sin h}{h}\right) \to 0 \times \sin x + 1 \times \cos x$ | dM1 | Dependent on both B and M marks. Complete attempt to apply limits: must isolate $\left(\frac{\cos h -1}{h}\right)$ and replace with 0, and isolate $\left(\frac{\sin h}{h}\right)$ and replace with 1. |
| so $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \cos x$ * | A1* | Printed answer — must be fully and correctly justified |12.
$$y = \sin x$$
where $x$ is measured in radians.\\
Use differentiation from first principles to show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \cos x$$
You may
\begin{itemize}
\item use without proof the formula for $\sin ( A \pm B )$
\item assume that as $h \rightarrow 0 , \frac { \sin h } { h } \rightarrow 1$ and $\frac { \cos h - 1 } { h } \rightarrow 0$
\end{itemize}
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q12 [5]}}