What is the derivative of $ \dfrac{{\cos x}}{{1 - \sin x}} $ ?

Hint: First of all we will apply the quotient rule identity and then will simplify for the resultant required value. Apply identity of the quotient rule as $ \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}} $

Complete step-by-step answer:
Take the given expression: $ y = \dfrac{{\cos x}}{{1 - \sin x}} $
Take the derivative on the both sides of the equation –
 $ \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{\cos x}}{{1 - \sin x}}} \right) $
Apply the identity formula in the above equation and then apply $ \left( {\dfrac{u}{v}} \right) $ rule.
Use the identity - $ \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}} $
 $ \dfrac{{dy}}{{dx}} = \dfrac{{(1 - \sin x)\dfrac{{d(\cos x)}}{{dx}} - \cos x\dfrac{d}{{dx}}(1 - \sin x)}}{{{{(1 - \sin x)}^2}}} $
Also, place that the derivative of the constant term is always zero
 $ \dfrac{{dy}}{{dx}} = \dfrac{{(1 - \sin x)( - \sin x) - \cos x( - \cos x)}}{{{{(1 - \sin x)}^2}}} $
Simplify the above expression –
 $ \dfrac{{dy}}{{dx}} = \dfrac{{ - \sin x + {{\sin }^2}x + {{\cos }^2}x}}{{{{(1 - \sin x)}^2}}} $
Use the identity $ {\cos ^2}x + {\sin ^2}x = 1 $
 $ \dfrac{{dy}}{{dx}} = \dfrac{{1 - \sin x}}{{{{(1 - \sin x)}^2}}} $
Common factors from the numerator and the denominator cancel each other.
 $ \dfrac{{dy}}{{dx}} = \dfrac{1}{{(1 - \sin x)}} $
This is the required solution.
So, the correct answer is “ $ \dfrac{1}{{(1 - \sin x)}} $ ”.

Note: Know the difference between the differentiation and the integration and apply formula accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverses of each other.