Question

If \[ y = \frac{\cos x}{1 + \sin x} \] then:

• A. \( \frac{dy}{dx} = \frac{-1}{1 + \sin x} \)
• B. \( \frac{dy}{dx} = \frac{1}{1 + \sin x} \)
• C. \( \frac{dy}{dx} = -\frac{1}{2} \sec^2 \left( \frac{\pi}{4} - \frac{x}{2} \right) \)
• D. \( \frac{dy}{dx} = -\frac{1}{2} \sec^2 \left( \frac{\pi}{4} - \frac{x}{2} \right) \)

Hint:

When differentiating a quotient, use the quotient rule \( \frac{dy}{dx} = \frac{(v \cdot u' - u \cdot v')}{v^2} \).

Answer 1

The Correct Option is B

We are given: \[ y = \frac{\cos x}{1 + \sin x} \] Now, to find \( \frac{dy}{dx} \), we will use the quotient rule of differentiation, which states: \[ \frac{dy}{dx} = \frac{(v \cdot u' - u \cdot v')}{v^2} \] Where \( u = \cos x \) and \( v = 1 + \sin x \), so their derivatives are: \( u' = -\sin x \) and \( v' = \cos x \). Applying the quotient rule: \[ \frac{dy}{dx} = \frac{(1 + \sin x)(-\sin x) - \cos x (\cos x)}{(1 + \sin x)^2} \] Simplifying: \[ \frac{dy}{dx} = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2} \] Now using the identity \( \sin^2 x + \cos^2 x = 1 \): \[ \frac{dy}{dx} = \frac{-\sin x - 1}{(1 + \sin x)^2} = \frac{-1}{1 + \sin x} \] Thus, the correct answer is option (b) \( \frac{dy}{dx} = \frac{-1}{1 + \sin x} \).