| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of quotient |
| Difficulty | Moderate -0.3 This is a standard C3 differentiation question testing routine application of chain rule, product rule, and quotient rule. Part (i) involves implicit differentiation with trig functions, (ii) is straightforward product rule with logarithms, and (iii) is quotient rule application. All parts follow textbook patterns with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07q 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 | Mark | Guidance |
| \(\frac{dx}{dy} = 4\sec^2 2y \tan 2y\) | B1 | Accept equivalent forms e.g. \(\frac{dx}{dy} = 4\frac{\sin 2y \cos 2y}{\cos^4 2y}\) |
| Use \(\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}\) | M1 | May be scored after next M1 if \(\frac{dx}{dy}\) written in terms of \(x\) |
| Uses \(\tan^2 2y = \sec^2 2y - 1\) and \(x = \sec^2 2y\) to get \(\frac{dx}{dy}\) or \(\frac{dy}{dx}\) in terms of just \(x\) | M1 | Do not accept \(y\)'s going to \(x\)'s directly e.g. \(\frac{dx}{dy} = 2\sec^2 2x\tan 2x \Rightarrow \frac{dy}{dx} = \frac{1}{2\sec^2 2x\tan 2x}\) scores M0 |
| \(\frac{dy}{dx} = \frac{1}{4x(x-1)^{\frac{1}{2}}}\) (conclusion stated with no errors previously) | A1* | 'Show that' question — all elements must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = (x^2+x^3)\times\frac{2}{2x} + (2x+3x^2)\ln 2x\) | M1 A1 A1 | M1: product rule applied. If rule stated must be correct. Only accept forms \(\ln 2x \times(ax+bx^2)+(x^2+x^3)\times\frac{C}{x}\) |
| When \(x=\frac{e}{2}\): \(\frac{dy}{dx} = 3\left(\frac{e}{2}\right) + 4\left(\frac{e}{2}\right)^2 = 3\left(\frac{e}{2}\right) + e^2\) | dM1 A1 | dM1: fully substitutes \(x=\frac{e}{2}\). Accept equivalent simplified forms: \(\frac{3e}{2}+e^2\), \(e(1.5+e)\), \(\frac{e(2e+3)}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = \frac{(x+1)^{\frac{1}{3}}(-3\sin x) - 3\cos x(\frac{1}{3}(x+1)^{-\frac{2}{3}})}{(x+1)^{\frac{2}{3}}}\) | M1 A1 | M1: quotient rule with \(u=3\cos x\), \(v=(x+1)^{\frac{1}{3}}\), \(u'=\pm A\sin x\), \(v'=B(x+1)^{-\frac{2}{3}}\). Also accept product rule approach |
| \(f'(x) = \frac{-3(x+1)(\sin x)-\cos x}{(x+1)^{\frac{4}{3}}}\) | A1 | Accept statement that \(g(x) = -3(x+1)(\sin x)-\cos x\) |
# Question 4(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dy} = 4\sec^2 2y \tan 2y$ | B1 | Accept equivalent forms e.g. $\frac{dx}{dy} = 4\frac{\sin 2y \cos 2y}{\cos^4 2y}$ |
| Use $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ | M1 | May be scored after next M1 if $\frac{dx}{dy}$ written in terms of $x$ |
| Uses $\tan^2 2y = \sec^2 2y - 1$ **and** $x = \sec^2 2y$ to get $\frac{dx}{dy}$ or $\frac{dy}{dx}$ in terms of just $x$ | M1 | Do not accept $y$'s going to $x$'s directly e.g. $\frac{dx}{dy} = 2\sec^2 2x\tan 2x \Rightarrow \frac{dy}{dx} = \frac{1}{2\sec^2 2x\tan 2x}$ scores M0 |
| $\frac{dy}{dx} = \frac{1}{4x(x-1)^{\frac{1}{2}}}$ (conclusion stated with no errors previously) | A1* | 'Show that' question — all elements must be seen |
# Question 4(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = (x^2+x^3)\times\frac{2}{2x} + (2x+3x^2)\ln 2x$ | M1 A1 A1 | M1: product rule applied. If rule stated must be correct. Only accept forms $\ln 2x \times(ax+bx^2)+(x^2+x^3)\times\frac{C}{x}$ |
| When $x=\frac{e}{2}$: $\frac{dy}{dx} = 3\left(\frac{e}{2}\right) + 4\left(\frac{e}{2}\right)^2 = 3\left(\frac{e}{2}\right) + e^2$ | dM1 A1 | dM1: fully substitutes $x=\frac{e}{2}$. Accept equivalent simplified forms: $\frac{3e}{2}+e^2$, $e(1.5+e)$, $\frac{e(2e+3)}{2}$ |
# Question 4(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = \frac{(x+1)^{\frac{1}{3}}(-3\sin x) - 3\cos x(\frac{1}{3}(x+1)^{-\frac{2}{3}})}{(x+1)^{\frac{2}{3}}}$ | M1 A1 | M1: quotient rule with $u=3\cos x$, $v=(x+1)^{\frac{1}{3}}$, $u'=\pm A\sin x$, $v'=B(x+1)^{-\frac{2}{3}}$. Also accept product rule approach |
| $f'(x) = \frac{-3(x+1)(\sin x)-\cos x}{(x+1)^{\frac{4}{3}}}$ | A1 | Accept statement that $g(x) = -3(x+1)(\sin x)-\cos x$ |\begin{enumerate}
\item (i) Given that
\end{enumerate}
$$x = \sec ^ { 2 } 2 y , \quad 0 < y < \frac { \pi } { 4 }$$
show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 4 x \sqrt { ( x - 1 ) } }$$
(ii) Given that
$$y = \left( x ^ { 2 } + x ^ { 3 } \right) \ln 2 x$$
find the exact value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at $x = \frac { \mathrm { e } } { 2 }$, giving your answer in its simplest form.\\
(iii) Given that
$$f ( x ) = \frac { 3 \cos x } { ( x + 1 ) ^ { \frac { 1 } { 3 } } } , \quad x \neq - 1$$
show that
$$\mathrm { f } ^ { \prime } ( x ) = \frac { \mathrm { g } ( x ) } { ( x + 1 ) ^ { \frac { 4 } { 3 } } } , \quad x \neq - 1$$
where $\mathrm { g } ( x )$ is an expression to be found.\\
\hfill \mbox{\textit{Edexcel C3 2014 Q4 [12]}}