Question

find $\frac{dy}{dx}$ for the following function.
y = $\frac{5cos x}{1 - sin x}$
$\frac{dy}{dx}=square$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = 5\cos x$, so $u'=- 5\sin x$, and $v = 1-\sin x$, so $v'=-\cos x$.

Step2: Substitute $u$, $u'$, $v$, and $v'$ into the quotient - rule

$\frac{dy}{dx}=\frac{(-5\sin x)(1 - \sin x)-5\cos x(-\cos x)}{(1 - \sin x)^{2}}$.

Step3: Expand the numerator

\[

$$\begin{align*} (-5\sin x)(1 - \sin x)-5\cos x(-\cos x)&=-5\sin x + 5\sin^{2}x+5\cos^{2}x\\ &=-5\sin x+5(\sin^{2}x+\cos^{2}x) \end{align*}$$

\]
Since $\sin^{2}x+\cos^{2}x = 1$, the numerator becomes $-5\sin x + 5$.

Step4: Simplify the derivative

$\frac{dy}{dx}=\frac{-5\sin x + 5}{(1 - \sin x)^{2}}=\frac{5(1-\sin x)}{(1 - \sin x)^{2}}=\frac{5}{1 - \sin x}$.

Answer:

$\frac{5}{1 - \sin x}$