find \\( \\frac { d y } { d x } \\) for the following function.\n\\( y = \\frac { 2 \\cos x } { 1 - \\sin x…

Explanation:

Step1: Apply the quotient rule

The quotient rule states that if (y = \frac{u}{v}), then (\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}). Here, (u = 2\cos x), (u'=- 2\sin x), (v = 1-\sin x), (v'=-\cos x).

Step2: Substitute into the quotient rule formula

[ \begin{align*} \frac{dy}{dx}&=\frac{(-2\sin x)(1 - \sin x)-2\cos x(-\cos x)}{(1-\sin x)^{2}}\ &=\frac{-2\sin x + 2\sin^{2}x + 2\cos^{2}x}{(1-\sin x)^{2}} \end{align*} ]

Step3: Use the trigonometric identity (\sin^{2}x+\cos^{2}x = 1)

[ \begin{align*} \frac{dy}{dx}&=\frac{-2\sin x+2(\sin^{2}x+\cos^{2}x)}{(1 - \sin x)^{2}}\ &=\frac{-2\sin x + 2\times1}{(1-\sin x)^{2}}\ &=\frac{2(1-\sin x)}{(1-\sin x)^{2}} \end{align*} ]

Step4: Simplify the expression

Cancel out the common factor ((1 - \sin x)) in the numerator and denominator (assuming (1-\sin x\neq0)), we get (\frac{2}{1-\sin x})

Answer:

(\frac{2}{1-\sin x})