Найдите производную третьего порядка. f(x)=1/x+sinx

Решите задачу:

$f(x)=\frac{1}{x+sinx}\; \; ,\quad (\frac{1}u})'=-\frac{u'}{u^2}\\\\f'(x)=-\frac{(x+sinx)'}{(x+sinx)^2}=-\frac{1+cosx}{(x+sinx)^2}\\\\f''(x)=-\frac{-sinx(x+sinx)^2-(1+cosx)\cdot 2(x+sinx)\cdot (1+cosx)}{(x+sinx)^4}=\frac{sinx\cdot (x+sinx)+2\cdot (1+cosx)^2}{(x+sinx)^3}\\\\f'''(x)=\frac{1}{(x+sinx)^3}\cdot \Big [\Big (cosx(x+sinx)+sinx(1+cosx)-\\\\-4(1+cosx)\cdot sinx\Big )(x+sinx)^3-\Big (sinx\cdot (x+sinx)+2\cdot (1+cosx)^2\Big )\times \\\\\times 3(x+sinx)^2\cdot (1+cosx)\Big ]$