Edexcel FP3 — Question 5 10 marks

Exam BoardEdexcel
ModuleFP3 (Further Pure Mathematics 3)
Marks10
PaperDownload PDF ↗
TopicHyperbolic functions
TypeDifferentiate inverse hyperbolic functions
DifficultyStandard +0.8 Part (a) is a standard inverse function differentiation using the chain rule (routine for FP3). Part (b) requires integration by parts with an inverse trig function, followed by careful algebraic manipulation and evaluation at non-trivial limits involving π/6. The 6-mark allocation and multi-step nature with exact form answer elevates this above average difficulty, but it follows standard FP3 techniques without requiring novel insight.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08i Integration by parts

  1. Given that \(y = \arctan 3x\), and assuming the derivative of \(\tan x\), prove that $$\frac{dy}{dx} = \frac{3}{1 + 9x^2}.$$ [4]
  2. Show that $$\int_0^{\frac{\sqrt{3}}{3}} 6x \arctan 3x \, dx = \frac{1}{3}(4\pi - 3\sqrt{3}).$$ [6]

\begin{enumerate}[label=(\alph*)]
\item Given that $y = \arctan 3x$, and assuming the derivative of $\tan x$, prove that
$$\frac{dy}{dx} = \frac{3}{1 + 9x^2}.$$ [4]

\item Show that
$$\int_0^{\frac{\sqrt{3}}{3}} 6x \arctan 3x \, dx = \frac{1}{3}(4\pi - 3\sqrt{3}).$$ [6]
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP3  Q5 [10]}}