Problem Statement
Let $f(x) = \ln(1 + \sin x)$ for $-\frac{\pi}{2} < x < \frac{\pi}{2}$.
**(a)** [3 marks] Show that $f'(x) = \frac{\cos x}{1 + \sin x}$ and $f''(x) = -\frac{1}{1 + \sin x}$.
**(b)** [3 marks] Find $f'''(0)$ and deduce the Maclaurin series expansion of $f(x)$ up to and including the term in $x^3$.
**(c)** [3 marks] Hence, evaluate $\lim_{x \to 0} \frac{\ln(1 + \sin x) - x}{x^2}$.
Verified Solution & Marking Scheme
Part (a) [M1]
Chain rule: $f'(x) = \frac{1}{1+\sin x} \cdot \cos x = \frac{\cos x}{1+\sin x}$
Part (a) [M1]
Quotient rule: $f''(x) = \frac{-\sin x(1+\sin x) - \cos x(\cos x)}{(1+\sin x)^2} = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2}$
Part (a) [A1]
$f''(x) = \frac{-(1+\sin x)}{(1+\sin x)^2} = -\frac{1}{1+\sin x}$ (AG)
Part (b) [M1]
$f'''(x) = \frac{d}{dx}[-(1+\sin x)^{-1}] = (1+\sin x)^{-2}\cos x = \frac{\cos x}{(1+\sin x)^2}$
Part (b) [A1]
$f(0) = \ln 1 = 0$, $f'(0) = 1$, $f''(0) = -1$, $f'''(0) = 1$
Part (b) [A1]
$f(x) = 0 + 1x + \frac{-1}{2!}x^2 + \frac{1}{3!}x^3 + \dots = x - \frac{1}{2}x^2 + \frac{1}{6}x^3$
Part (c) [M1]
Substitute series: $\frac{(x - \frac{1}{2}x^2 + \frac{1}{6}x^3 + \dots) - x}{x^2}$
Part (c) [A1]
$= \frac{-\frac{1}{2}x^2 + \frac{1}{6}x^3}{x^2} = -\frac{1}{2} + \frac{1}{6}x$
Part (c) [A1]
Taking limit as $x \to 0$: $= -\frac{1}{2}$