C3 January 2010 Q4
4.
| Scheme | Marks |
|---|---|
| \(y = \dfrac{\ln(x^2 + 1)}{x}\) | |
| \(u = \ln(x^2 + 1) \Rightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{2x}{x^2 + 1}\) | M1 A1 |
| Apply quotient rule: \(\left\{\begin{aligned} u &= \ln(x^2 + 1) & v &= x \\ \frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{2x}{x^2 + 1} & \frac{\mathrm{d}v}{\mathrm{d}x} &= 1 \end{aligned}\right\}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\left(\frac{2x}{x^2 + 1}\right)(x) - \ln(x^2 + 1)}{x^2}\) | M1 A1 |
| \(\left\{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{(x^2 + 1)} - \dfrac{1}{x^2}\ln(x^2 + 1)\right\}\) | |
| (4) |
Notes
M1: \(\ln(x^2 + 1) \to \dfrac{\text{something}}{x^2 + 1}\)
A1: \(\ln(x^2 + 1) \to \dfrac{2x}{x^2 + 1}\)
M1: Applying \(\dfrac{xu' - \ln(x^2 + 1)v'}{x^2}\) correctly.
A1: Correct differentiation with correct bracketing but allow recovery.
{Ignore subsequent working.}
| Scheme | Marks |
|---|---|
| \(x = \tan y\) | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \sec^2 y\) | M1* A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sec^2 y}\ \{= \cos^2 y\}\) | dM1* |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 + \tan^2 y}\) | dM1* |
| Hence, \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 + x^2}\), (as required) | A1 AG |
| (5) | |
| (9 marks) |
Notes
M1*: \(\tan y \to \sec^2 y\) or an attempt to differentiate \(\dfrac{\sin y}{\cos y}\) using either the quotient rule or product rule.
A1: \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \sec^2 y\)
dM1*: Finding \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) by reciprocating \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\).
dM1*: For writing down or applying the identity \(\sec^2 y = 1 + \tan^2 y\), which must be applied/stated completely in \(y\).
A1 AG: For the correct proof, leading on from the previous line of working.