FP3 June 2014 Q5
5. Given that \(y = \mathrm{artanh}\dfrac{x}{\sqrt{(1 + x^2)}}\)
show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{(1 + x^2)}}\) (4)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \underline{\left(\dfrac{1}{1 - \dfrac{x^2}{1 + x^2}}\right)} \cdot \underline{\underline{\left(\dfrac{(1 + x^2)^{\frac{1}{2}} - x^2(1 + x^2)^{-\frac{1}{2}}}{(1 + x^2)}\right)}}\) NB \(\dfrac{(1 + x^2)^{\frac{1}{2}} - x^2(1 + x^2)^{-\frac{1}{2}}}{(1 + x^2)} = \dfrac{1}{(1 + x^2)^{\frac{3}{2}}}\) M1: Correct form for the derivative of \(\mathrm{artanh}\,x\) using \(\dfrac{x}{\sqrt{1 + x^2}}\). M1: Correct quotient or product rule on \(\dfrac{x}{\sqrt{1 + x^2}}\) A1: Completely correct expression | M1M1 A1 |
| \(= \dfrac{1}{\sqrt{1 + x^2}}\) **ag** Correct solution with no errors seen | A1 |
| (4) | |
| (4 marks) |
Notes
Alternative 1
| Scheme | Marks |
|---|---|
| \(y = \mathrm{artanh}\dfrac{x}{\sqrt{1 + x^2}} \Rightarrow \tanh y = \dfrac{x}{\sqrt{1 + x^2}}\) \(\mathrm{sech}^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \left(\dfrac{(1 + x^2)^{\frac{1}{2}} - x^2(1 + x^2)^{-\frac{1}{2}}}{(1 + x^2)}\right)\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \underline{\left(\dfrac{1}{1 - \dfrac{x^2}{1 + x^2}}\right)} \cdot \underline{\underline{\left(\dfrac{(1 + x^2)^{\frac{1}{2}} - x^2(1 + x^2)^{-\frac{1}{2}}}{(1 + x^2)}\right)}}\) M1: Divides by the correct form of \(\mathrm{sech}^2 y\) or their simplified \(\mathrm{sech}^2 y\) in terms of \(x\) M1: Correct quotient or product rule A1: Completely correct expression | M1M1 A1 |
| Then as above |
Alternative 2
| Scheme | Marks |
|---|---|
| \(y = \mathrm{artanh}\dfrac{x}{\sqrt{1 + x^2}} \Rightarrow \tanh y = \dfrac{x}{\sqrt{1 + x^2}} \Rightarrow \tanh^2 y = \dfrac{x^2}{1 + x^2}\) \(2\tanh y\,\mathrm{sech}^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \left(\dfrac{2x(1 + x^2) - 2x^3}{(1 + x^2)^2}\right)\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \underline{\left(\dfrac{1}{1 - \dfrac{x^2}{1 + x^2}}\right) \times \dfrac{(1 + x^2)^{\frac{1}{2}}}{2x}} \times \underline{\underline{\dfrac{2x(1 + x^2) - 2x^3}{(1 + x^2)^2}}}\) M1: Divides by the correct form of \(\mathrm{sech}^2 y\) and \(\tanh y\) in terms of \(x\) M1: Correct quotient or product rule A1: Completely correct expression | M1M1 A1 |
| Then as above |