FP3 June 2017 Q1
1. Given that \(y = \text{arsinh}(\tanh x)\), show that \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\text{sech}^2 x}{\sqrt{1 + \tanh^2 x}}\] (5)
| Scheme | Marks |
|---|---|
| \(y = \text{arsinh}(\tanh x)\) | |
| Way 1: \(\sinh y = \tanh x\) | B1 |
| \(\cosh y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \text{sech}^2 x\) or \(\cosh y = \text{sech}^2 x\dfrac{\mathrm{d}x}{\mathrm{d}y}\) | M1A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\text{sech}^2 x}{\cosh y}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\text{sech}^2 x}{\sqrt{1 + \sinh^2 y}} = \mathrm{f}(x)\) | M1 |
| \(= \dfrac{\text{sech}^2 x}{\sqrt{1 + \tanh^2 x}}\) * | A1* |
| (5 marks) |
Notes
B1: \(\sinh y = \tanh x\)
M1: \(\pm\cosh y\) or \(\pm\text{sech}^2 x\)
A1: All correct
M1: Uses a correct identity to express \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) only
A1*: cso. There must be no errors such as incorrect or missing or inconsistent variables and no missing h’s.
Way 2
| Scheme | Marks |
|---|---|
| \(t = \tanh x \Rightarrow y = \text{arsinh}\,t\) | B1 |
| \(\dfrac{\mathrm{d}t}{\mathrm{d}x} = \text{sech}^2 x,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{1}{\sqrt{1 + t^2}}\) | M1A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t}\dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{\text{sech}^2 x}{\sqrt{1 + t^2}} = \mathrm{f}(x)\) | M1 |
| \(= \dfrac{\text{sech}^2 x}{\sqrt{1 + \tanh^2 x}}\) * | A1* |
| (5 marks) |
B1: Replaces \(\tanh x\) by e.g. \(t\)
M1: \(\dfrac{\mathrm{d}t}{\mathrm{d}x} = \pm\text{sech}^2 x,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = \pm\dfrac{1}{\sqrt{1 + t^2}}\)
A1: Correct \(\dfrac{\mathrm{d}t}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) and correctly labelled
M1: Uses correct form of the chain rule for their variables to express \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) only
A1*: Cso. There must be no errors such as incorrect or missing or inconsistent variables and no missing h’s.
Way 3
| Scheme | Marks |
|---|---|
| \(u = \tanh x \Rightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x} = \text{sech}^2 x\) | B1 |
| \(\displaystyle\int\frac{\text{sech}^2 x}{\sqrt{1 + \tanh^2 x}}\,\mathrm{d}x = \int\frac{\text{sech}^2 x}{\sqrt{1 + u^2}}\frac{1}{\text{sech}^2 x}\,\mathrm{d}u\) | M1A1 |
| \(\displaystyle = \int\frac{1}{\sqrt{1 + u^2}}\,\mathrm{d}u = \text{arsinh}\,u\ (+c)\) | M1 |
| \(y = \text{arsinh}(\tanh x)(+c)\) | A1* |
| (5 marks) |
B1: Correct derivative
M1: Complete substitution including the “\(\mathrm{d}x\)”
A1: Fully correct substitution
M1: Reaches \(\text{arsinh}\,u\)
A1*: Reaches \(y = \text{arsinh}(\tanh x)\) with or without + c and no errors such as incorrect or missing or inconsistent variables or missing h’s.
Special Case
| Scheme | Marks |
|---|---|
| \(y = \text{arsinh}(\tanh x) \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{1 + \tanh^2 x}}(\times)\,\text{sech}^2 x\) \(= \dfrac{\text{sech}^2 x}{\sqrt{1 + \tanh^2 x}}\) Note that the sech2\(x\) needs to appear separate from the fraction as above and not just the printed answer written down. To score more than 2 marks using a chain rule method, a third variable must be introduced | M1A1 |