| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of product |
| Difficulty | Moderate -0.8 This is a straightforward multi-part differentiation question testing standard rules (chain rule for parts a and c, product rule for part b). Each part requires direct application of a single technique with minimal algebraic manipulation. This is easier than average as it's purely procedural with no problem-solving or insight required. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07l Derivative of ln(x): and related functions1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks |
|---|---|
| \(= \frac{-1}{\cos x} \times (-\sin x) = -\tan x\) | M1 A2 |
| Answer | Marks |
|---|---|
| \(= 2x \times \sin 3x + x^2 \times 3\cos 3x = 2x \sin 3x + 3x^2 \cos 3x\) | M1 A2 |
| Answer | Marks |
|---|---|
| \(= \frac{d}{dx}[6(2x-7)^{-\frac{1}{2}}]\) | B1 |
| \(= -3(2x-7)^{-\frac{3}{2}} \times 2 = -\frac{6}{(2x-7)^{\frac{3}{2}}}\) | M1 A2 |
| (10) |
**Part (a)**
$= \frac{-1}{\cos x} \times (-\sin x) = -\tan x$ | M1 A2 |
**Part (b)**
$= 2x \times \sin 3x + x^2 \times 3\cos 3x = 2x \sin 3x + 3x^2 \cos 3x$ | M1 A2 |
**Part (c)**
$= \frac{d}{dx}[6(2x-7)^{-\frac{1}{2}}]$ | B1 |
$= -3(2x-7)^{-\frac{3}{2}} \times 2 = -\frac{6}{(2x-7)^{\frac{3}{2}}}$ | M1 A2 |
| (10) |
---3. Differentiate each of the following with respect to $x$ and simplify your answers.
\begin{enumerate}[label=(\alph*)]
\item $\quad \ln ( \cos x )$
\item $x ^ { 2 } \sin 3 x$
\item $\frac { 6 } { \sqrt { 2 x - 7 } }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q3 [10]}}