The derivative of $\dfrac1{1 + \sin x}$ is $\boxed{-\dfrac{\cos x}{(1 + \sin x)^2}}$.
To find the derivative of $\dfrac1{1 + \sin x}$, we will use a combination of the power rule and chain rule. The power rule is:
\[\begin{align*} \dfrac{d}{dx} x^n &= nx^{n – 1} \end{align*}\]
The chain rule is:
\[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)) \end{align*}\]
Notice that $\dfrac1{1 + \sin x}$ may be rewritten as $(1 + \sin x)^{-1}$. Here, our two functions are $f(x) = \dfrac1x, g(x) = 1 + \sin x$. With this in mind, we may proceed as follows:
\[\begin{align*} \dfrac{d}{dx} \dfrac1{1 + \sin x} &= \dfrac{d}{dx} (1 + \sin x)^{-1}\\ &= -\frac{1}{\left(1+\sin x\right)^2}\frac{d}{dx}\left(1+\sin x\right)\\ &= -\frac{1}{\left(1+\sin x\right)^2}\cos x\\ &= -\dfrac{\cos x}{(1 + \sin x)^2} \end{align*}\]
We get that the derivative of $\dfrac1{1 + \sin x}$ is $\boxed{-\dfrac{\cos x}{(1 + \sin x)^2}}$.
For more complex derivatives, it is handy to break down the derivative into separate functions that are easier to handle.
In this case, we took $f(x) = \dfrac1x, g(x) = 1 + \sin x$, which is a common tactic. It is normal to treat $\dfrac1x$ as its own function in instances where fractions are involved.
