find \\( \\frac { d y } { d x } \\) for the following function.\n\n\\( y = \\frac { 6 \\cos x } { 1 - \\sin…

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 = 6\cos x), (u'=- 6\sin x), (v = 1-\sin x), (v'=-\cos x). [ \begin{align*} \frac{dy}{dx}&=\frac{(-6\sin x)(1 - \sin x)-6\cos x(-\cos x)}{(1-\sin x)^{2}}\ \end{align*} ]

Step2: Expand the numerator

Expand ((-6\sin x)(1 - \sin x)-6\cos x(-\cos x)): [ \begin{align*} (-6\sin x)(1 - \sin x)-6\cos x(-\cos x)&=-6\sin x + 6\sin^{2}x+6\cos^{2}x\ \end{align*} ]

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

Substitute (\sin^{2}x+\cos^{2}x = 1) into the numerator: [ \begin{align*} -6\sin x + 6\sin^{2}x+6\cos^{2}x&=-6\sin x+6(\sin^{2}x+\cos^{2}x)\ &=-6\sin x + 6\times1\ &=6-6\sin x\ &=6(1 - \sin x) \end{align*} ]

Step4: Simplify the derivative

[ \begin{align*} \frac{dy}{dx}&=\frac{6(1 - \sin x)}{(1-\sin x)^{2}}\ &=\frac{6}{1-\sin x} \end{align*} ]

Answer:

(\frac{6}{1-\sin x})