4.
$$y = \operatorname { artanh } \left( \frac { \cos x + a } { \cos x - a } \right)$$
where \(a\) is a non-zero constant.
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Question 4:
\[y = \text{artanh}\left(\frac{\cos x + a}{\cos x - a}\right)\]
Answer Marks
Guidance
Working/Answer Mark
Guidance
\(\frac{dy}{dx} = \frac{1}{1-\left(\frac{\cos x+a}{\cos x-a}\right)^2} \times \frac{(\cos x-a)\times -\sin x - (\cos x+a)\times -\sin x}{(\cos x-a)^2}\) M1
Correct method for derivative. Requires \(\frac{1}{1-\left(\frac{\cos x+a}{\cos x-a}\right)^2}\) × attempt at quotient (or product) rule
Correct derivative in any form A1
\(= \frac{(\cos x-a)^2}{(\cos x-a)^2-(\cos x+a)^2} \times \frac{2a\sin x}{(\cos x-a)^2} = \frac{2a\sin x}{-4a\cos x} = \ldots\) dM1
Uses correct processing to reach \(\lambda\frac{\sin x}{\cos x}\) or \(\lambda\tan x\). Depends on first method mark
\(= -\frac{1}{2}\tan x\) A1
cso
Way 2:
Answer Marks
Guidance
Working/Answer Mark
Guidance
\(\tanh y = \frac{\cos x+a}{\cos x-a} \Rightarrow \text{sech}^2 y\frac{dy}{dx} = \frac{2a\sin x}{(\cos x-a)^2}\) M1
Takes tanh of both sides, obtains \(\text{sech}^2 y\frac{dy}{dx}\) = attempt at quotient or product rule
\(\frac{dy}{dx} = \frac{1}{1-\left(\frac{\cos x+a}{\cos x-a}\right)^2} \times \frac{2a\sin x}{(\cos x-a)^2}\) A1
Correct derivative in any form
Uses correct processing to reach \(\lambda\frac{\sin x}{\cos x}\) or \(\lambda\tan x\) dM1
Depends on first method mark
\(= -\frac{1}{2}\tan x\) A1
cso
Way 3:
Answer Marks
Guidance
Working/Answer Mark
Guidance
\(u = \frac{\cos x+a}{\cos x-a}\), obtains \(\frac{du}{dx} = \frac{2a\sin x}{(\cos x-a)^2}\) by quotient rule and \(\frac{dy}{du} = \frac{1}{1-u^2}\), chain rule gives \(\frac{dy}{dx} = \frac{1}{1-\left(\frac{\cos x+a}{\cos x-a}\right)^2} \times \frac{2a\sin x}{(\cos x-a)^2}\) M1
Correct derivative in any form A1
Uses correct processing to reach \(\lambda\frac{\sin x}{\cos x}\) or \(\lambda\tan x\) dM1
Depends on first method mark
\(= -\frac{1}{2}\tan x\) A1
cso
Way 4:
Answer Marks
Guidance
Working/Answer Mark
Guidance
\(y = \frac{1}{2}\ln\left(\frac{1+\frac{\cos x+a}{\cos x-a}}{1-\frac{\cos x+a}{\cos x-a}}\right) = \frac{1}{2}\ln\left(-\frac{\cos x}{a}\right)\), \(\frac{dy}{dx} = \frac{1}{2}\times\frac{1}{-\frac{\cos x}{a}}\times\left(\frac{\sin x}{a}\right)\) M1A1
M1: Converts to correct ln form and uses chain rule to differentiate. A1: Correct derivative in any form
Uses correct processing to reach \(\lambda\frac{\sin x}{\cos x}\) or \(\lambda\tan x\) dM1
Depends on first method mark
\(= -\frac{1}{2}\tan x\) A1
cso
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# Question 4:
$$y = \text{artanh}\left(\frac{\cos x + a}{\cos x - a}\right)$$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{1-\left(\frac{\cos x+a}{\cos x-a}\right)^2} \times \frac{(\cos x-a)\times -\sin x - (\cos x+a)\times -\sin x}{(\cos x-a)^2}$ | M1 | Correct method for derivative. Requires $\frac{1}{1-\left(\frac{\cos x+a}{\cos x-a}\right)^2}$ × attempt at quotient (or product) rule |
| Correct derivative in any form | A1 | |
| $= \frac{(\cos x-a)^2}{(\cos x-a)^2-(\cos x+a)^2} \times \frac{2a\sin x}{(\cos x-a)^2} = \frac{2a\sin x}{-4a\cos x} = \ldots$ | dM1 | Uses correct processing to reach $\lambda\frac{\sin x}{\cos x}$ or $\lambda\tan x$. Depends on first method mark |
| $= -\frac{1}{2}\tan x$ | A1 | cso |
**Way 2:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\tanh y = \frac{\cos x+a}{\cos x-a} \Rightarrow \text{sech}^2 y\frac{dy}{dx} = \frac{2a\sin x}{(\cos x-a)^2}$ | M1 | Takes tanh of both sides, obtains $\text{sech}^2 y\frac{dy}{dx}$ = attempt at quotient or product rule |
| $\frac{dy}{dx} = \frac{1}{1-\left(\frac{\cos x+a}{\cos x-a}\right)^2} \times \frac{2a\sin x}{(\cos x-a)^2}$ | A1 | Correct derivative in any form |
| Uses correct processing to reach $\lambda\frac{\sin x}{\cos x}$ or $\lambda\tan x$ | dM1 | Depends on first method mark |
| $= -\frac{1}{2}\tan x$ | A1 | cso |
**Way 3:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u = \frac{\cos x+a}{\cos x-a}$, obtains $\frac{du}{dx} = \frac{2a\sin x}{(\cos x-a)^2}$ by quotient rule and $\frac{dy}{du} = \frac{1}{1-u^2}$, chain rule gives $\frac{dy}{dx} = \frac{1}{1-\left(\frac{\cos x+a}{\cos x-a}\right)^2} \times \frac{2a\sin x}{(\cos x-a)^2}$ | M1 | |
| Correct derivative in any form | A1 | |
| Uses correct processing to reach $\lambda\frac{\sin x}{\cos x}$ or $\lambda\tan x$ | dM1 | Depends on first method mark |
| $= -\frac{1}{2}\tan x$ | A1 | cso |
**Way 4:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y = \frac{1}{2}\ln\left(\frac{1+\frac{\cos x+a}{\cos x-a}}{1-\frac{\cos x+a}{\cos x-a}}\right) = \frac{1}{2}\ln\left(-\frac{\cos x}{a}\right)$, $\frac{dy}{dx} = \frac{1}{2}\times\frac{1}{-\frac{\cos x}{a}}\times\left(\frac{\sin x}{a}\right)$ | M1A1 | M1: Converts to correct ln form and uses chain rule to differentiate. A1: Correct derivative in any form |
| Uses correct processing to reach $\lambda\frac{\sin x}{\cos x}$ or $\lambda\tan x$ | dM1 | Depends on first method mark |
| $= -\frac{1}{2}\tan x$ | A1 | cso |
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4.
$$y = \operatorname { artanh } \left( \frac { \cos x + a } { \cos x - a } \right)$$
where $a$ is a non-zero constant.\\
Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = k \tan x$$
where $k$ is a constant to be determined.
\hfill \mbox{\textit{Edexcel F3 2022 Q4 [4]}}