Question 9 - A-Level Maths

Edexcel PMT Mocks — Question 9 5 marks

Exam Board	Edexcel
Module	PMT Mocks (PMT Mocks)
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 careful algebraic manipulation of sin(x+h), application of given limit results, and multi-step reasoning. While the structure is well-scaffolded with provided limits, it demands more sophistication than routine differentiation and is typically challenging for A-level students encountering rigorous proof.
Spec	1.07h Differentiation from first principles: for sin(x) and cos(x)

9. Given that $x$ is measured in radians, prove, from the first principles, that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( \sin x ) = \cos x$$ You may assume the formula for $\sin ( A \pm B )$ and that as $h \rightarrow 0 , \frac { \sin h } { h } \rightarrow 1$ and $\frac { \cos h - 1 } { h } \rightarrow 0$.

Show mark scheme Show mark scheme source

Question 9: Prove from first principles that $\frac{d}{dx}(\sin x) = \cos x$

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$f'(x) = \lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$	B1	Correct fraction stated
Uses $\sin(x+h) = \sin x\cos h + \cos x\sin h$	M1	Compound angle formula applied
Achieves $\frac{\sin x\cos h + \cos x\sin h}{h}$	A1	Correct expanded numerator over $h$
Complete attempt to apply given limits $\frac{\cos h - 1}{h}\to 0$ and $\frac{\sin h}{h}\to 1$	M1	Both limits applied to gradient expression
Correct conclusion: $\frac{d}{dx}(\sin x) = \cos x$	A1	Fully correct solution only

(5 marks)

## Question 9: Prove from first principles that $\frac{d}{dx}(\sin x) = \cos x$

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = \lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$ | B1 | Correct fraction stated |
| Uses $\sin(x+h) = \sin x\cos h + \cos x\sin h$ | M1 | Compound angle formula applied |
| Achieves $\frac{\sin x\cos h + \cos x\sin h}{h}$ | A1 | Correct expanded numerator over $h$ |
| Complete attempt to apply given limits $\frac{\cos h - 1}{h}\to 0$ and $\frac{\sin h}{h}\to 1$ | M1 | Both limits applied to gradient expression |
| Correct conclusion: $\frac{d}{dx}(\sin x) = \cos x$ | A1 | Fully correct solution only |

**(5 marks)**

Show LaTeX source

9. Given that $x$ is measured in radians, prove, from the first principles, that

$$\frac { \mathrm { d } } { \mathrm {~d} x } ( \sin x ) = \cos x$$

You may assume the formula for $\sin ( A \pm B )$ and that as $h \rightarrow 0 , \frac { \sin h } { h } \rightarrow 1$ and $\frac { \cos h - 1 } { h } \rightarrow 0$.\\

\hfill \mbox{\textit{Edexcel PMT Mocks  Q9 [5]}}

This paper (14 questions)

View full paper

Q1 6 Q2 5 Q3 5 Q4 7 Q5 5 Q6 7 Q7 9 Q8 7 Q9 5 Q10 6 Q11 7 Q12 7 Q13 10 Q14 14

Edexcel PMT Mocks — Question 9 5 marks

Question 9: Prove from first principles that \(\frac{d}{dx}(\sin x) = \cos x\)

(5 marks)

This paper (14 questions)