Question

find \\(\frac{dy}{dx}\\) for \\(y = \frac{\sec x}{1 + \sec x}\\)

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

Explanation:

Simplify the function using trigonometric identities

$$ y = \frac{\sec x}{1 + \sec x} = \frac{\frac{1}{\cos x}}{1 + \frac{1}{\cos x}} = \frac{1}{\cos x + 1} = (\cos x + 1)^{-1} $$

Apply the chain rule to find the derivative

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

Simplify the derivative expression

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

Answer:

Find \(\frac{dy}{dx}\) for \(y = \frac{\sec x}{1 + \sec x}\).

\(\frac{dy}{dx} =\) <blank>\(\frac{\sin x}{(1 + \cos x)^2}\)</blank>