find $ \frac { d y } { d x } $ for the following function.

$ y = \frac { 6 \cos x } { 1 - \sin x } $

$ \frac { d y } { d x } = $

Question

find $ \frac { d y } { d x } $ for the following function.

$ y = \frac { 6 \cos x } { 1 - \sin x } $

$ \frac { d y } { d x } = $

Accepted Answer

# 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:
$\frac{6}{1-\sin x}$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply the quotient rule

Step2: Expand the numerator

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

Step4: Simplify the derivative

Answer: