| Exam Board | WJEC |
|---|---|
| Module | Unit 3 (Unit 3) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | First principles: trigonometric functions |
| Difficulty | Standard +0.3 Part (a) is a standard first principles proof requiring knowledge of the limit of (sin h)/h, worth 5 marks but following a well-established method. Parts (b)(i) and (b)(ii) are routine applications of quotient rule and product rule respectively. While the first principles derivation requires careful working, it's a textbook exercise that students are explicitly taught, and the subsequent parts are straightforward differentiation practice. |
| Spec | 1.07h Differentiation from first principles: for sin(x) and cos(x)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07q Product and quotient rules: differentiation |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = \lim_{h \to 0} \left[\frac{\cos(x+h) - \cos x}{h}\right]\) | M1 |
| \(= \lim_{h \to 0} \left[\frac{\cos x \cos h - \sin x \sin h - \cos x}{h}\right]\) | A1 |
| As \(h\) approaches \(0\): \(\cos h \approx 1 - \frac{h^2}{2}\) and \(\sin h \approx h\) | |
| So \(\frac{dy}{dx} = \lim_{h \to 0} \left[\frac{\cos x\left(1-\frac{h^2}{2}\right) - \sin x \times h - \cos x}{h}\right]\) | M1 |
| \(= \lim_{h \to 0} \left[\frac{-\frac{h^2}{2}\cos x - h\sin x}{h}\right]\) | A1 |
| \(= -\sin x\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{(x^3+1)6x - 3x^2(3x^2)}{(x^3+1)^2}\) | M1 | (Correct formula) |
| \(= \frac{3x(2-x^3)}{(x^3+1)^2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(3x^2 \tan 3x + 3x^3 \sec^2 3x\) | M1 | (Correct formula) (All Correct) |
| \(= 3x^2(\tan 3x + x\sec^2 3x)\) | A1 |
### Part (a):
Let $y = \cos x$
$\frac{dy}{dx} = \lim_{h \to 0} \left[\frac{\cos(x+h) - \cos x}{h}\right]$ | M1 |
$= \lim_{h \to 0} \left[\frac{\cos x \cos h - \sin x \sin h - \cos x}{h}\right]$ | A1 |
As $h$ approaches $0$: $\cos h \approx 1 - \frac{h^2}{2}$ and $\sin h \approx h$ | |
So $\frac{dy}{dx} = \lim_{h \to 0} \left[\frac{\cos x\left(1-\frac{h^2}{2}\right) - \sin x \times h - \cos x}{h}\right]$ | M1 |
$= \lim_{h \to 0} \left[\frac{-\frac{h^2}{2}\cos x - h\sin x}{h}\right]$ | A1 |
$= -\sin x$ | A1 |
### Part (b)(i):
$\frac{(x^3+1)6x - 3x^2(3x^2)}{(x^3+1)^2}$ | M1 | (Correct formula)
$= \frac{3x(2-x^3)}{(x^3+1)^2}$ | A1 |
### Part (b)(ii):
$3x^2 \tan 3x + 3x^3 \sec^2 3x$ | M1 | (Correct formula) (All Correct)
$= 3x^2(\tan 3x + x\sec^2 3x)$ | A1 |
**Total: [9]**\begin{enumerate}[label=(\alph*)]
\item Differentiate $\cos x$ from first principles. [5]
\item Differentiate the following with respect to $x$, simplifying your answer as far as possible.
\begin{enumerate}[label=(\roman*)]
\item $\frac{3x^2}{x^3+1}$ [2]
\item $x^3 \tan 3x$ [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 3 Q12 [9]}}