First Few Terms in a Maclaurin Series

Question

This is the question

I found the derivatives of ln(1+sinx) and I keep getting $x$ $-$ $x^2$$/2$ $+$ $x^3$$/3$ $-$ $x^4$$/4$ as my terms. None of the answers have those so I'm not sure what I'm doing wrong here.

Answer 1

To get the series for composed functions up to a small degree, you can compose series. Let's see, $$ \log(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \; \mbox{more} $$

in the above, set $$ t \approx \sin x = x - \frac{x^3}{6} + \; \mbox{more} $$

So, just put $t = x - \frac{x^3}{6}$ into $t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4}$ and keep only terms with $x$ exponent up to 4

Answer 2

$$ f(x) = \ln (1+\sin x) \implies f(0)=0$$

$$f'(x) = \frac {\cos x}{1+\sin x} \implies f'(0) = 1$$

$$ f''(x) = \frac { -\sin x (1+\sin x)-\cos^2 x}{(1+\sin x )^2} \implies f''(0) = -1$$

Continue and you get

$$ \ln( 1 + \sin x ) = x-x^2/2 + x^3/6 -...$$

Which is the choice (e)