Problem Statement
Consider the function $f(x) = \ln(1 + \sin x)$ for $-\frac{\pi}{2} < x < \frac{\pi}{2}$.
**(a)** [4 marks] Find $f(0), f'(0), f''(0),$ and $f'''(0)$.
**(b)** [3 marks] Hence, state the Maclaurin series expansion for $f(x)$ up to and including the term in $x^3$.
Verified Solution & Marking Scheme
Part (a) [A1]
$f(0) = \ln(1 + \sin 0) = \ln 1 = 0$
Part (a) [M1]
$f'(x) = \frac{\cos x}{1 + \sin x} \implies f'(0) = \frac{\cos 0}{1 + \sin 0} = \frac{1}{1} = 1$
Part (a) [M1]
$f''(x) = \frac{-\sin x(1 + \sin x) - \cos x(\cos x)}{(1 + \sin x)^2} = \frac{-\sin x - (\sin^2 x + \cos^2 x)}{(1 + \sin x)^2} = \frac{-(1 + \sin x)}{(1 + \sin x)^2} = -\frac{1}{1 + \sin x} \implies f''(0) = -1$
Part (a) [A1]
$f'''(x) = \frac{d}{dx}[-(1 + \sin x)^{-1}] = (1 + \sin x)^{-2}\cos x \implies f'''(0) = \frac{\cos 0}{(1 + \sin 0)^2} = 1$
Part (b) [M1]
Recall formula: $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$
Part (b) [A1]
$f(x) = 0 + 1(x) + \frac{-1}{2}x^2 + \frac{1}{6}x^3 + \dots$
Part (b) [A1]
$f(x) = x - \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$