Given that \(y = \cos\{\ln(1 + x)\}\), prove that
- \((1 + x)\frac{\mathrm{d}y}{\mathrm{d}x} = -\sin\{\ln(1 + x)\}\), [1]
- \((1 + x)^2 \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (1 + x)\frac{\mathrm{d}y}{\mathrm{d}x} + y = 0\). [2]
Obtain an equation relating \(\frac{\mathrm{d}^3 y}{\mathrm{d}x^3}\), \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\) and \(\frac{\mathrm{d}y}{\mathrm{d}x}\). [2]
Hence find Maclaurin's series for \(y\), up to and including the term in \(x^3\). [4]
Verify that the same result is obtained if the standard series expansions for \(\ln(1 + x)\) and \(\cos x\) are used. [3]