2. Determine the second derivative \( \frac{d^{\wedge} 2 y}{d x^{\wedge} 2} \) of \( \sec 2 \mathrm{x} \). ( 10 marks)

To determine the second derivative of \( y = \sec(2x) \), we first need to find the first derivative. 1. **First Derivative**: Using the chain rule, the derivative of \( \sec(u) \) is \( \sec(u) \tan(u) \frac{du}{dx} \). Here, \( u = 2x \), so: \[ \frac{dy}{dx} = \sec(2x) \tan(2x) \cdot 2 = 2 \sec(2x) \tan(2x) \] 2. **Second Derivative**: Next, we will differentiate \( \frac{dy}{dx} = 2 \sec(2x) \tan(2x) \) again using the product rule. Let \( u = \sec(2x) \) and \( v = \tan(2x) \): \[ \frac{d^2y}{dx^2} = 2 \left( \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx} \right) \] Now we need to compute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \): - **Calculating \( \frac{du}{dx} \)**: \[ \frac{du}{dx} = 2 \sec(2x) \tan(2x) \] - **Calculating \( \frac{dv}{dx} \)**: \[ \frac{dv}{dx} = 2 \sec^2(2x) \] Now substituting back: \[ \frac{d^2y}{dx^2} = 2 \left( (2 \sec(2x) \tan(2x)) \tan(2x) + \sec(2x) (2 \sec^2(2x)) \right) \] \[ = 4 \sec(2x) \tan^2(2x) + 4 \sec^3(2x) \] Thus, the second derivative is: \[ \frac{d^2y}{dx^2} = 4 \sec(2x) \tan^2(2x) + 4 \sec^3(2x) \]