| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Implicit or inverse differentiation |
| Difficulty | Standard +0.3 This is a structured, multi-part question that guides students through standard techniques: quotient rule application, trigonometric identities, inverse function differentiation, and finding/classifying stationary points. While it requires multiple steps and connects several concepts, each individual part uses routine C3 methods with clear signposting, making it slightly easier than average. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07e Second derivative: as rate of change of gradient1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
9
\begin{enumerate}[label=(\alph*)]
\item Given that $x = \frac { \sin y } { \cos y }$, use the quotient rule to show that
$$\frac { \mathrm { d } x } { \mathrm {~d} y } = \sec ^ { 2 } y$$
(3 marks)
\item Given that $\tan y = x - 1$, use a trigonometrical identity to show that
$$\sec ^ { 2 } y = x ^ { 2 } - 2 x + 2$$
\item Show that, if $y = \tan ^ { - 1 } ( x - 1 )$, then
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { x ^ { 2 } - 2 x + 2 }$$
(l mark)
\item A curve has equation $y = \tan ^ { - 1 } ( x - 1 ) - \ln x$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of the $x$-coordinate of each of the stationary points of the curve.
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item Hence show that the curve has a minimum point which lies on the $x$-axis.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2012 Q9 [14]}}