What is the derivative of \[f(x) = \dfrac{{\cos x}}{{1 + \sin x}}?\]

Hint: We have to find the differentiation of the function \[\dfrac{{\cos x}}{{1 + \sin x}}\] . Take $\cos x$ as the first function and $1 + \sin x$ as the second function and differentiate by using the quotient rule. The rule which is to be used to solve this question is
$\dfrac{d}{{dx}}\left( {\dfrac{{p(x)}}{{q(x)}}} \right) = \dfrac{{q(x)p'(x) - p(x)q'(x)}}{{{{(q(x))}^2}}}$ , where $p(x)$ is the first function and $q(x)$ is the second function.

Complete step by step solution:
From the question, we are given the function \[\dfrac{{\cos x}}{{1 + \sin x}}\] . In this function we can clearly see that this function is of the two differentiable functions. So, we can use the quotient rule of differentiation to solve this question. In calculus, the quotient rule can be defined as the method for finding the differentiation of that function which is the ratio of two differentiable functions. Let $y = \dfrac{{p(x)}}{{q(x)}}$ , where both the functions $p$ and $q$ are differentiable and $q(x) \ne 0$ . Then , by using the quotient rule the derivative of the given function can be done as-
$\dfrac{d}{{dx}}\left( {\dfrac{{p(x)}}{{q(x)}}} \right) = \dfrac{{q(x)p'(x) - p(x)q'(x)}}{{{{(q(x))}^2}}}$
We have the function \[\dfrac{{\cos x}}{{1 + \sin x}}\] .
Therefore, here $f(x) = \dfrac{{\cos x}}{{1 + \sin x}}$ , $p(x) = \cos x$ and $q(x) = 1 + \sin x$
Putting the values of $f(x),p(x)$ and $q(x)$ from the above in the equation , we get
$ \Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{{(1 + \sin x)\dfrac{d}{{dx}}(\cos x) - (\cos x)\dfrac{d}{{dx}}(1 + \sin x)}}{{{{(1 + \sin x)}^2}}}$
We know that the differentiation of $\cos x$ is $( - \sin x)$ and the differentiation of $(1 + \sin x)$ is $\cos x$ .
Put this in the above equation and we get
$ \Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{{(1 + \sin x)( - \sin x) - (\cos x)(\cos x)}}{{{{(1 + \sin x)}^2}}}$
$ \Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{{ - \sin x - {{\sin }^2}x - {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}}$
$ \Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{{ - \sin x - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{(1 + \sin x)}^2}}}$
We know that ${\sin ^2}x + {\cos ^2}x = 1$ , put this in above equation and we get
$ \Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{{ - \sin x - 1}}{{{{(1 + \sin x)}^2}}}$
$ \Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{{ - (1 + \sin x)}}{{{{(1 + \sin x)}^2}}}$
Omitting the same term and we get
$ \Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{{ - 1}}{{1 + \sin x}}$
Hence, $\dfrac{{ - 1}}{{1 + \sin x}}$ is the required derivative of the function of \[\dfrac{{\cos x}}{{1 + \sin x}}\] .

Note:
There are six trigonometric functions. These are sine, cosine, tangent, cosecant, secant and cotangent. The property of the trigonometry we use in the above solution is \[{\sin ^2}x + {\cos ^2}x = 1\] . Always try to simplify your answer as much as you can because it gives a good impression on your answer.