| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Moderate -0.3 This is a standard C3 differentiation question testing product rule (part a), quotient rule (part b), and second derivative (part c). All parts follow routine procedures with no novel insight required. Part (b) being a 'show that' makes it slightly easier than open-ended questions, and part (c) is straightforward algebra after finding the second derivative. Slightly easier than average due to its procedural nature. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07j Differentiate exponentials: e^(kx) and a^(kx)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 | Guidance |
|---|---|---|
| (a)(i) \(\frac{d}{dx}(e^x(\sin x + 2\cos x)) = 3e^x(\sin x + 2\cos x) + e^x(\cos x - 2\sin x)\) | M1 A1 A1 | |
| \(= e^x(\sin x + 7\cos x)\) | (3) | |
| (ii) \(\frac{d}{dx}(x^3\ln(5x+2)) = 3x^2\ln(5x+2) + \frac{5x^3}{5x+2}\) | M1 A1 A1 | (3) |
| (b) \(\frac{dy}{dx} = \frac{(x+1)^2(6x+6) - 2(x+1)(3x^2+6x-7)}{(x+1)^4}\) | M1 A1 A1 | |
| \(= \frac{(x+1)(6x^2+12x+6-6x^2-12x+14)}{(x+1)^4}\) | M1 | |
| \(= \frac{20}{(x+1)^3}\) | cso A1 | (5) |
| (c) \(\frac{d^2y}{dx^2} = -\frac{60}{(x+1)^4} = -\frac{15}{4}\) | M1 | |
| \((x+1)^4 = 16\) | M1 | |
| \(x = 1, -3\) both | A1 | (3) [14] |
| Answer | Marks |
|---|---|
| \(\frac{(x+1)^2(6x+6) - 2(x+1)(3x^2+6x-7)}{(x+1)^4} = \frac{(6x^3+18x^2+18x+6)-(6x^3+18x^2-2x-14)}{(x+1)^4} = \frac{20x+20}{(x+1)^4} = \frac{20(x+1)}{(x+1)^4} = \frac{20}{(x+1)^3}\) | M1 A1 |
**(a)(i)** $\frac{d}{dx}(e^x(\sin x + 2\cos x)) = 3e^x(\sin x + 2\cos x) + e^x(\cos x - 2\sin x)$ | M1 A1 A1 |
$= e^x(\sin x + 7\cos x)$ | (3)
**(ii)** $\frac{d}{dx}(x^3\ln(5x+2)) = 3x^2\ln(5x+2) + \frac{5x^3}{5x+2}$ | M1 A1 A1 | (3)
**(b)** $\frac{dy}{dx} = \frac{(x+1)^2(6x+6) - 2(x+1)(3x^2+6x-7)}{(x+1)^4}$ | M1 A1 A1 |
$= \frac{(x+1)(6x^2+12x+6-6x^2-12x+14)}{(x+1)^4}$ | M1 |
$= \frac{20}{(x+1)^3}$ | cso A1 | (5)
**(c)** $\frac{d^2y}{dx^2} = -\frac{60}{(x+1)^4} = -\frac{15}{4}$ | M1 |
$(x+1)^4 = 16$ | M1 |
$x = 1, -3$ both | A1 | (3) [14]
**Note:** The simplification in part (b) can be carried out as:
$\frac{(x+1)^2(6x+6) - 2(x+1)(3x^2+6x-7)}{(x+1)^4} = \frac{(6x^3+18x^2+18x+6)-(6x^3+18x^2-2x-14)}{(x+1)^4} = \frac{20x+20}{(x+1)^4} = \frac{20(x+1)}{(x+1)^4} = \frac{20}{(x+1)^3}$ | M1 A1 |
---6. (a) Differentiate with respect to $x$,
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { e } ^ { 3 x } ( \sin x + 2 \cos x )$,
\item $x ^ { 3 } \ln ( 5 x + 2 )$.
Given that $y = \frac { 3 x ^ { 2 } + 6 x - 7 } { ( x + 1 ) ^ { 2 } } , \quad x \neq - 1$,\\
(b) show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 20 } { ( x + 1 ) ^ { 3 } }$.\\
(c) Hence find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ and the real values of $x$ for which $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - \frac { 15 } { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2008 Q6 [14]}}