Solution

(a) First Derivative:

Differentiate implicitly with respect to \( x \):

\[ \frac{dy}{dx} – \sin y \frac{dy}{dx} = 1 \] \[ \frac{dy}{dx} (1 – \sin y) = 1 \] \[ \frac{dy}{dx} = \frac{1}{1 – \sin y} \]

(b) Vertical Tangents:

Vertical tangents occur when \( \frac{dy}{dx} \) is undefined:

\[ 1 – \sin y = 0 \Rightarrow \sin y = 1 \] \[ y = \frac{\pi}{2} \text{ (within given domain)} \]

Substitute into original equation:

\[ \frac{\pi}{2} + \cos\left(\frac{\pi}{2}\right) = x + 1 \] \[ \frac{\pi}{2} + 0 = x + 1 \Rightarrow x = \frac{\pi}{2} – 1 \]

Equation of vertical tangent: \( x = \frac{\pi}{2} – 1 \)

(c) Second Derivative:

Differentiate \( \frac{dy}{dx} \) using chain rule:

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{1 – \sin y}\right) = \frac{-\left(-\cos y \frac{dy}{dx}\right)}{(1 – \sin y)^2} \] \[ = \frac{\cos y \left(\frac{1}{1 – \sin y}\right)}{(1 – \sin y)^2} = \frac{\cos y}{(1 – \sin y)^3} \]