| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Differentiate inverse trigonometric functions |
| Difficulty | Standard +0.8 This FP3 question requires product rule with arcsin (standard but less routine than basic calculus), exact evaluation involving surds, and a non-trivial algebraic manipulation to express the derivative of arctan(3e^{2x}) in hyperbolic form. Part (b) particularly demands recognizing how to convert 1+9e^{4x} into the given hyperbolic expression, which requires insight beyond mechanical differentiation. Moderately challenging for Further Maths. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08g Derivatives: inverse trig and hyperbolic functions |
\begin{enumerate}[label=(\alph*)]
\item Given that $y = x \arcsin x$, $0 \leq x \leq 1$, find
\begin{enumerate}[label=(\roman*)]
\item an expression for $\frac{dy}{dx}$,
\item the exact value of $\frac{dy}{dx}$ when $x = \frac{1}{2}$.
\end{enumerate}
[3]
\item Given that $y = \arctan(3e^{2x})$, show that
$$\frac{dy}{dx} = \frac{3}{5\cosh 2x + 4\sinh 2x}.$$
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2011 Q2 [8]}}