| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Differentiate inverse trigonometric functions |
| Difficulty | Standard +0.3 Part (a) is a standard bookwork proof of the derivative of arcsin(x) using implicit differentiation - a routine Core Pure 2 exercise. Part (b) requires applying the chain rule to find f'(x) and showing it's never zero, which is straightforward once part (a) is established. This is slightly easier than average as it's mostly direct application of learned techniques with minimal problem-solving. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin y = x \Rightarrow \cos y \frac{dy}{dx} = 1\) (or \(\frac{dx}{dy} = \cos y\)) | M1 | Finds \(x\) in terms of \(y\) and differentiates |
| Uses \(\sin^2 y + \cos^2 y = 1 \Rightarrow \cos y = \sqrt{1-\sin^2 y} \Rightarrow \sqrt{1-x^2}\) | M1 | Uses trig identity to express \(\cos y\) in terms of \(x\) |
| \(\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\) | A1* | Correctly achieves printed answer, cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = \frac{1}{\sqrt{1-e^{2x}}} \times \ldots\) (chain rule applied) | M1 | Differentiates using chain rule to achieve correct form; condone \(f'(x) = \frac{1}{\sqrt{1-e^{2x}}}\); note \(\frac{1}{\sqrt{1-e^x}}\) is B0 |
| \(f'(x) = \frac{1}{\sqrt{1-e^{2x}}} \times e^x = \frac{e^x}{\sqrt{1-e^{2x}}}\) | A1 | Correct differentiation |
| \(e^x \neq 0\) (or \(e^x > 0\)) therefore no stationary points; or \(e^x = 0 \Rightarrow x = \ln 0\) which is impossible/undefined | A1 | Follows correct differentiation; states \(e^x \neq 0\) or no solutions to \(e^x = 0\); ignore reference to denominator \(= 0\) |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin y = x \Rightarrow \cos y \frac{dy}{dx} = 1$ (or $\frac{dx}{dy} = \cos y$) | M1 | Finds $x$ in terms of $y$ and differentiates |
| Uses $\sin^2 y + \cos^2 y = 1 \Rightarrow \cos y = \sqrt{1-\sin^2 y} \Rightarrow \sqrt{1-x^2}$ | M1 | Uses trig identity to express $\cos y$ in terms of $x$ |
| $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$ | A1* | Correctly achieves printed answer, cso |
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## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = \frac{1}{\sqrt{1-e^{2x}}} \times \ldots$ (chain rule applied) | M1 | Differentiates using chain rule to achieve correct form; condone $f'(x) = \frac{1}{\sqrt{1-e^{2x}}}$; note $\frac{1}{\sqrt{1-e^x}}$ is B0 |
| $f'(x) = \frac{1}{\sqrt{1-e^{2x}}} \times e^x = \frac{e^x}{\sqrt{1-e^{2x}}}$ | A1 | Correct differentiation |
| $e^x \neq 0$ (or $e^x > 0$) therefore no stationary points; or $e^x = 0 \Rightarrow x = \ln 0$ which is impossible/undefined | A1 | Follows correct differentiation; states $e^x \neq 0$ or no solutions to $e^x = 0$; ignore reference to denominator $= 0$ |\begin{enumerate}
\item (a) Given that
\end{enumerate}
$$y = \arcsin x \quad - 1 \leqslant x \leqslant 1$$
show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$$
(b)
$$\mathrm { f } ( x ) = \arcsin \left( \mathrm { e } ^ { x } \right) \quad x \leqslant 0$$
Prove that $\mathrm { f } ( x )$ has no stationary points.
\hfill \mbox{\textit{Edexcel CP2 2022 Q5 [6]}}