Question

find \\(\frac{dy}{dx}\\) for the following function.

\\y = \frac{3 \cos x}{1 + \sin x}\\

\\(\frac{dy}{dx} = \square\\)

Explanation:

Apply the quotient rule

$$ \frac{dy}{dx} = \frac{\frac{d}{dx}[3\cos x] \cdot (1 + \sin x) - 3\cos x \cdot \frac{d}{dx}[1 + \sin x]}{(1 + \sin x)^2} $$

Differentiate the numerator and denominator terms

$$ \frac{dy}{dx} = \frac{-3\sin x(1 + \sin x) - 3\cos x(\cos x)}{(1 + \sin x)^2} $$

Simplify the expression

$$
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Answer:

Find \(\frac{dy}{dx}\) for the following function.

$$y = \frac{3\cos x}{1 + \sin x}$$

\(\frac{dy}{dx} =\) <blank>\(\frac{-3}{1 + \sin x}\)</blank>