| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Differentiation of logarithmic functions |
| Difficulty | Moderate -0.8 This is a straightforward application of standard differentiation rules. Part (a) requires the chain rule for ln(f(x)), giving f'(x)/f(x), which is routine. Part (b) uses the quotient rule (or product rule with x^{-2}), also standard. Both are direct applications of learned techniques with no problem-solving or insight required, making this easier than average. |
| Spec | 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 | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{(x^2+3x+5)} \times \ldots = \frac{2x+3}{(x^2+3x+5)}\) | M1, A1 | |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Applying \(\frac{vu'-uv'}{v^2}\) | M1 | |
| \(\frac{x^2 \times -\sin x - \cos x \times 2x}{(x^2)^2} = \frac{-x^2\sin x - 2x\cos x}{x^4} = \frac{-x\sin x - 2\cos x}{x^3}\) | A2,1,0 | oe |
| (3) | 5 Marks |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{(x^2+3x+5)} \times \ldots = \frac{2x+3}{(x^2+3x+5)}$ | M1, A1 | |
| | (2) | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Applying $\frac{vu'-uv'}{v^2}$ | M1 | |
| $\frac{x^2 \times -\sin x - \cos x \times 2x}{(x^2)^2} = \frac{-x^2\sin x - 2x\cos x}{x^4} = \frac{-x\sin x - 2\cos x}{x^3}$ | A2,1,0 | oe |
| | (3) | **5 Marks** |
---Differentiate with respect to $x$
\begin{enumerate}[label=(\alph*)]
\item $\quad \ln \left( x ^ { 2 } + 3 x + 5 \right)$
\item $\frac { \cos x } { x ^ { 2 } }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2011 Q1 [5]}}