Quotient rule differentiation and evaluation

With $f(x)=\cos(x)$ and $g(x)=1+\sin(x)$, we have $f'(x) = -\sin(x)$ and $g'(x) = \cos(x)$. Applying the quotient rule: $h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} = \frac{-\sin(x)(1+\sin(x)) - \cos(x)\cos(x)}{(1+\sin(x))^2}.$

Expanding the numerator gives $-\sin(x) - \sin^2(x) - \cos^2(x) = -\sin(x) - 1$ (using $\sin^2x+\cos^2x=1$). So $h'(x) = \dfrac{-\sin(x)-1}{(1+\sin(x))^2} = \dfrac{-1}{1+\sin(x)}$. Evaluating at $x=0$: $\sin(0)=0$, so $h'(0) = \dfrac{-1}{1+0} = -1$.

The simplified derivative is negative one divided by one plus sine of x. Since $\sin(x)$ ranges from $-1$ to $1$, the expression $1 + \sin(x)$ ranges from $0$ to $2$. The function $h(x)$ and its derivative $h'(x)$ are undefined when $1 + \sin(x) = 0$, which occurs when $\sin(x) = -1$ (at $x = -\frac{\pi}{2} + 2\pi k$ for integer $k$). Wherever $h'(x)$ is defined (i.e., where $1 + \sin(x) \neq 0$), we have $1 + \sin(x) > 0$, which means the numerator is negative and the denominator is positive. Therefore, $h'(x)$ is always negative wherever it is defined.