| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Inverse function differentiation |
| Difficulty | Standard +0.8 This is a Further Maths question requiring knowledge of inverse trig differentiation and integration by parts with arctan. Part (a) needs chain rule with the standard arctan derivative, while part (b) requires recognizing arctan as the function to keep when applying integration by parts—a non-routine technique that students often find challenging. The algebraic simplification in both parts adds moderate complexity. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{2}{3} \cdot \frac{1}{1+\frac{4x^2}{9}}\) | M1 | Use formula for derivative of arctan: \(\frac{dy}{dx} = \frac{p}{1+(qx)^2}\), \(q \neq 1\). Condone missing brackets around \(qx\) but must be \(1+(qx)^2\) not \(1-(qx)^2\), \(p\) may be 1 |
| \(= \frac{6}{9+4x^2}\) | A1 | Answer as shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \arctan\left(\frac{2x}{3}\right)dx = \left[x\arctan\left(\frac{2x}{3}\right)\right] - \int \frac{6x}{9+4x^2}dx\) | M1A1ft | M1: Use of parts in correct direction. Allow e.g. \(x\arctan\left(\frac{2x}{3}\right) - \int x\,d\!\left(\arctan\left(\frac{2x}{3}\right)\right)\). A1ft: Follow through from part (a) |
| \(= \left[x\arctan\left(\frac{2x}{3}\right)\right] - \left[\frac{3}{4}\ln(9+4x^2)\right](+c)\) | M1A1 | M1: Use of ln correctly for their fraction. A1: cao (+ c not required). Allow \(x\arctan\left(\frac{2x}{3}\right)\times x\) and \(-\frac{3}{4}\ln k(9+4x^2)\) |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{2}{3} \cdot \frac{1}{1+\frac{4x^2}{9}}$ | M1 | Use formula for derivative of arctan: $\frac{dy}{dx} = \frac{p}{1+(qx)^2}$, $q \neq 1$. Condone missing brackets around $qx$ but must be $1+(qx)^2$ not $1-(qx)^2$, $p$ may be 1 |
| $= \frac{6}{9+4x^2}$ | A1 | Answer **as shown** |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \arctan\left(\frac{2x}{3}\right)dx = \left[x\arctan\left(\frac{2x}{3}\right)\right] - \int \frac{6x}{9+4x^2}dx$ | M1A1ft | M1: Use of parts in correct direction. Allow e.g. $x\arctan\left(\frac{2x}{3}\right) - \int x\,d\!\left(\arctan\left(\frac{2x}{3}\right)\right)$. A1ft: Follow through from part (a) |
| $= \left[x\arctan\left(\frac{2x}{3}\right)\right] - \left[\frac{3}{4}\ln(9+4x^2)\right](+c)$ | M1A1 | M1: Use of ln correctly for their fraction. A1: cao (+ c not required). Allow $x\arctan\left(\frac{2x}{3}\right)\times x$ and $-\frac{3}{4}\ln k(9+4x^2)$ |
---\begin{enumerate}
\item Given that $y = \arctan \left( \frac { 2 x } { 3 } \right)$,\\
(a) find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving your answer in its simplest form.\\
(b) Use integration by parts to find
\end{enumerate}
$$\int \arctan \left( \frac { 2 x } { 3 } \right) \mathrm { d } x$$
\hfill \mbox{\textit{Edexcel F3 2014 Q1 [6]}}