| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| 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 techniques: chain rule for part (a), product rule for (b)(i), and finding a tangent equation in (b)(ii). All parts are routine textbook exercises requiring direct application of learned rules with no problem-solving or insight needed, making it easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 6(x^3-1)^5 \times 3x^2\) | M1 | Chain rule attempt |
| \(= 18x^2(x^3-1)^5\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = x \cdot \frac{1}{x} + \ln x\) | M1 | Product rule attempt |
| \(= 1 + \ln x\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(x = e\): \(y = e \ln e = e\) | B1 | Correct point \((e, e)\) |
| Gradient \(= 1 + \ln e = 2\) | M1 | Substituting \(x = e\) into their \(\frac{dy}{dx}\) |
| \(y - e = 2(x - e)\) | M1 | Line equation using their gradient and point |
| \(y = 2x - e\) | A1 | Correct simplified equation |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 6(x^3-1)^5 \times 3x^2$ | M1 | Chain rule attempt |
| $= 18x^2(x^3-1)^5$ | A1 | Correct final answer |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = x \cdot \frac{1}{x} + \ln x$ | M1 | Product rule attempt |
| $= 1 + \ln x$ | A1 | Correct final answer |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| When $x = e$: $y = e \ln e = e$ | B1 | Correct point $(e, e)$ |
| Gradient $= 1 + \ln e = 2$ | M1 | Substituting $x = e$ into their $\frac{dy}{dx}$ |
| $y - e = 2(x - e)$ | M1 | Line equation using their gradient and point |
| $y = 2x - e$ | A1 | Correct simplified equation |
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\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when $y = \left( x ^ { 3 } - 1 \right) ^ { 6 }$.
\item A curve has equation $y = x \ln x$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Find an equation of the tangent to the curve $y = x \ln x$ at the point on the curve where $x = \mathrm { e }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2011 Q1 [7]}}