| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Differentiate inverse trigonometric functions |
| Difficulty | Standard +0.8 Part (a) is a standard FP3 derivation using implicit differentiation, but part (b) requires careful application of the product rule and quotient rule to differentiate the result from (a), then algebraic manipulation to reach the given form. While systematic, this is more demanding than typical C3/C4 questions and requires multiple differentiation techniques in sequence. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07a Derivative as gradient: of tangent to curve1.07d Second derivatives: d^2y/dx^2 notation |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \arcsin x \Rightarrow \sin y = x\) | M1 | |
| \(\cos y \frac{dy}{dx} = 1\) | ||
| \(\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}\) | M1A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = -\frac{1}{2}(1-x^2)^{-\frac{3}{2}}(-2x)\) | M1A1 | |
| \(= x(1-x^2)^{-\frac{3}{2}}\) | ||
| \((1-x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = (1-x^2) \cdot x(1-x^2)^{-\frac{3}{2}} - x(1-x^2)^{-\frac{1}{2}} = 0\) | M1A1 | (4) |
## (a)
$y = \arcsin x \Rightarrow \sin y = x$ | M1 |
$\cos y \frac{dy}{dx} = 1$ | |
$\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}$ | M1A1 | (3)
## (b)
$\frac{d^2y}{dx^2} = -\frac{1}{2}(1-x^2)^{-\frac{3}{2}}(-2x)$ | M1A1 |
$= x(1-x^2)^{-\frac{3}{2}}$ | |
$(1-x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = (1-x^2) \cdot x(1-x^2)^{-\frac{3}{2}} - x(1-x^2)^{-\frac{1}{2}} = 0$ | M1A1 | (4)
---Given that $y = \arcsin x$ prove that
\begin{enumerate}[label=(\alph*)]
\item $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$ [3]
\item $(1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0$ [4]
\end{enumerate}
(Total 7 marks)
\hfill \mbox{\textit{Edexcel FP3 Q5 [7]}}