5 Given that \(y = \ln \left( \sqrt { \frac { 2 x - 1 } { 2 x + 1 } } \right)\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 x - 1 } - \frac { 1 } { 2 x + 1 }\).
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Question 5(1) – Additional Solutions:
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\(y = \ln\!\left(\sqrt{\frac{2x-1}{2x+1}}\right) = \frac{1}{2}\ln\!\left(\frac{2x-1}{2x+1}\right)\) \(\Rightarrow \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\left(\frac{2x-1}{2x+1}\right)} \cdot \frac{(2x+1)2-(2x-1)2}{(2x+1)^2}\) M1
\(\ln\sqrt{c} = \frac{1}{2}\ln c\) used
\(= \frac{1}{2}\cdot\frac{2x+1}{2x-1}\cdot\frac{4}{(2x+1)^2} = \frac{2}{(2x-1)(2x+1)}\) A2
fully correct expression for the derivative
\(\frac{1}{2x-1} - \frac{1}{2x+1} = \frac{2x+1-(2x-1)}{(2x-1)(2x+1)} = \frac{2}{(2x-1)(2x+1)}\) A1
simplified and shown to be equivalent to \(\frac{1}{2x-1} - \frac{1}{2x+1}\)
[4]
Question 5(2) – Additional Solutions:
Answer Marks
Guidance
Answer Marks
Guidance
\(y = \ln\!\left(\sqrt{\frac{2x-1}{2x+1}}\right) = \ln\sqrt{2x-1} - \ln\sqrt{2x+1}\) M1
\(\ln(a/b) = \ln a - \ln b\) used
\(\frac{dy}{dx} = \frac{1}{\sqrt{2x-1}}\cdot\frac{1}{2}(2x-1)^{-1/2}\cdot 2 - \frac{1}{\sqrt{2x+1}}\cdot\frac{1}{2}(2x+1)^{-1/2}\cdot 2\) A2
fully correct expression
\(= \frac{1}{2x-1} - \frac{1}{2x+1}\) A1
simplified and shown to be equivalent to \(\frac{1}{2x-1} - \frac{1}{2x+1}\)
[4]
Question 5(3) – Additional Solutions:
Answer Marks
Guidance
Answer Marks
Guidance
\(y = \ln\!\left(\sqrt{\frac{2x-1}{2x+1}}\right)\) \(\frac{dy}{dx} = \frac{1}{\sqrt{\frac{2x-1}{2x+1}}} \cdot \frac{1}{2}\!\left(\frac{2x-1}{2x+1}\right)^{-1/2} \cdot \frac{(2x+1)2-(2x-1)2}{(2x+1)^2}\) M1
\(\frac{1}{u}\times\) their \(u'\) where \(u = \sqrt{\frac{2x-1}{2x+1}}\) or \(\frac{\sqrt{2x-1}}{\sqrt{2x+1}}\) (any attempt at \(u'\) will do)
A2 o.e. any completely correct expression for the derivative
\(\frac{1}{2x-1} - \frac{1}{2x+1} = \frac{(2x+1)-(2x-1)}{(2x-1)(2x+1)} = \frac{2}{(2x-1)(2x+1)}\) A1
simplified and correctly shown to be equivalent to \(\frac{1}{2x-1} - \frac{1}{2x+1}\)
[4]
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## Question 5(1) – Additional Solutions:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \ln\!\left(\sqrt{\frac{2x-1}{2x+1}}\right) = \frac{1}{2}\ln\!\left(\frac{2x-1}{2x+1}\right)$ | | |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\left(\frac{2x-1}{2x+1}\right)} \cdot \frac{(2x+1)2-(2x-1)2}{(2x+1)^2}$ | M1 | $\ln\sqrt{c} = \frac{1}{2}\ln c$ used |
| $= \frac{1}{2}\cdot\frac{2x+1}{2x-1}\cdot\frac{4}{(2x+1)^2} = \frac{2}{(2x-1)(2x+1)}$ | A2 | fully correct expression for the derivative |
| $\frac{1}{2x-1} - \frac{1}{2x+1} = \frac{2x+1-(2x-1)}{(2x-1)(2x+1)} = \frac{2}{(2x-1)(2x+1)}$ | A1 | simplified and shown to be equivalent to $\frac{1}{2x-1} - \frac{1}{2x+1}$ |
| | **[4]** | |
## Question 5(2) – Additional Solutions:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \ln\!\left(\sqrt{\frac{2x-1}{2x+1}}\right) = \ln\sqrt{2x-1} - \ln\sqrt{2x+1}$ | M1 | $\ln(a/b) = \ln a - \ln b$ used |
| $\frac{dy}{dx} = \frac{1}{\sqrt{2x-1}}\cdot\frac{1}{2}(2x-1)^{-1/2}\cdot 2 - \frac{1}{\sqrt{2x+1}}\cdot\frac{1}{2}(2x+1)^{-1/2}\cdot 2$ | A2 | fully correct expression |
| $= \frac{1}{2x-1} - \frac{1}{2x+1}$ | A1 | simplified and shown to be equivalent to $\frac{1}{2x-1} - \frac{1}{2x+1}$ |
| | **[4]** | |
## Question 5(3) – Additional Solutions:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \ln\!\left(\sqrt{\frac{2x-1}{2x+1}}\right)$ | | |
| $\frac{dy}{dx} = \frac{1}{\sqrt{\frac{2x-1}{2x+1}}} \cdot \frac{1}{2}\!\left(\frac{2x-1}{2x+1}\right)^{-1/2} \cdot \frac{(2x+1)2-(2x-1)2}{(2x+1)^2}$ | M1 | $\frac{1}{u}\times$ their $u'$ where $u = \sqrt{\frac{2x-1}{2x+1}}$ or $\frac{\sqrt{2x-1}}{\sqrt{2x+1}}$ (any attempt at $u'$ will do) |
| | A2 | o.e. any completely correct expression for the derivative |
| $\frac{1}{2x-1} - \frac{1}{2x+1} = \frac{(2x+1)-(2x-1)}{(2x-1)(2x+1)} = \frac{2}{(2x-1)(2x+1)}$ | A1 | simplified and correctly shown to be equivalent to $\frac{1}{2x-1} - \frac{1}{2x+1}$ |
| | **[4]** | |
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5 Given that $y = \ln \left( \sqrt { \frac { 2 x - 1 } { 2 x + 1 } } \right)$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 x - 1 } - \frac { 1 } { 2 x + 1 }$.
\hfill \mbox{\textit{OCR MEI C3 2013 Q5 [4]}}