| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of quotient |
| Difficulty | Moderate -0.8 This is a straightforward multi-part differentiation question testing standard rules (chain rule for ln, quotient rule, and product rule with exponentials). Each part is routine application of a single technique with minimal algebraic manipulation required, making it easier than average for A-level. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07l Derivative of ln(x): and related functions1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(= \frac{1}{3x-2} \times 3 = \frac{3}{3x-2}\) | M1 A1 | |
| (b) \(= \frac{2x(1-x) - (2x+1)x(-1)}{(1-x)^2} = \frac{3}{(1-x)^2}\) | M1 A2 | |
| (c) \(= \frac{2}{3}x^{\frac{1}{3}}e^{2x} + x^{\frac{1}{3}} \times 2e^{2x} = \frac{1}{3}x^{\frac{1}{3}}e^{2x}(3+4x)\) | M1 A2 | (8 marks) |
**(a)** $= \frac{1}{3x-2} \times 3 = \frac{3}{3x-2}$ | M1 A1 |
**(b)** $= \frac{2x(1-x) - (2x+1)x(-1)}{(1-x)^2} = \frac{3}{(1-x)^2}$ | M1 A2 |
**(c)** $= \frac{2}{3}x^{\frac{1}{3}}e^{2x} + x^{\frac{1}{3}} \times 2e^{2x} = \frac{1}{3}x^{\frac{1}{3}}e^{2x}(3+4x)$ | M1 A2 | (8 marks)
---4. Differentiate each of the following with respect to $x$ and simplify your answers.
\begin{enumerate}[label=(\alph*)]
\item $\quad \ln ( 3 x - 2 )$
\item $\frac { 2 x + 1 } { 1 - x }$
\item $x ^ { \frac { 3 } { 2 } } \mathrm { e } ^ { 2 x }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q4 [8]}}