| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of quotient |
| Difficulty | Moderate -0.3 This is a straightforward application of standard differentiation rules (product rule, quotient rule, chain rule, and implicit differentiation). Part (a) requires direct application of formulas with no problem-solving, while part (b) is a routine implicit differentiation exercise. Slightly easier than average due to being purely procedural with no conceptual challenges. |
| Spec | 1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d}{dx}(e^{3x+2}) = 3e^{3x+2}\) | B1 | At any stage; or \(3e^2e^{3x}\) |
| \(\frac{dy}{dx} = 3x^2 e^{3x+2} + 2xe^{3x+2}\) | M1 A1+A1 | Or equivalent (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d}{dx}(\cos(2x^3)) = -6x^2\sin(2x^3)\) | M1 A1 | At any stage |
| \(\frac{dy}{dx} = \frac{-18x^3\sin(2x^3) - 3\cos(2x^3)}{9x^2}\) | M1 A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 = 8\cos(2y+6)\frac{dy}{dx}\) or \(\frac{dx}{dy} = 8\cos(2y+6)\) | M1 | |
| \(\frac{dy}{dx} = \frac{1}{8\cos(2y+6)}\) | M1 A1 | |
| \(\frac{dy}{dx} = \frac{1}{8\cos\left(\arcsin\left(\frac{x}{4}\right)\right)}\) \(\left(= \pm\frac{1}{2\sqrt{16-x^2}}\right)\) | M1 A1 | (5) [13] |
# Question 4:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}(e^{3x+2}) = 3e^{3x+2}$ | B1 | At any stage; or $3e^2e^{3x}$ |
| $\frac{dy}{dx} = 3x^2 e^{3x+2} + 2xe^{3x+2}$ | M1 A1+A1 | Or equivalent **(4)** |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}(\cos(2x^3)) = -6x^2\sin(2x^3)$ | M1 A1 | At any stage |
| $\frac{dy}{dx} = \frac{-18x^3\sin(2x^3) - 3\cos(2x^3)}{9x^2}$ | M1 A1 | **(4)** |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 = 8\cos(2y+6)\frac{dy}{dx}$ or $\frac{dx}{dy} = 8\cos(2y+6)$ | M1 | |
| $\frac{dy}{dx} = \frac{1}{8\cos(2y+6)}$ | M1 A1 | |
| $\frac{dy}{dx} = \frac{1}{8\cos\left(\arcsin\left(\frac{x}{4}\right)\right)}$ $\left(= \pm\frac{1}{2\sqrt{16-x^2}}\right)$ | M1 A1 | **(5) [13]** |
---4. (a) Differentiate with respect to $x$
\begin{enumerate}[label=(\roman*)]
\item $x ^ { 2 } \mathrm { e } ^ { 3 x + 2 }$,
\item $\frac { \cos \left( 2 x ^ { 3 } \right) } { 3 x }$.\\
(b) Given that $x = 4 \sin ( 2 y + 6 )$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2006 Q4 [13]}}