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

We are given the function   y = sec(2x). Step 1. Find the first derivative (dy/dx). Recall that the derivative of sec(u) with respect to x is sec(u)tan(u) · (du/dx). Here u = 2x, so du/dx = 2. Thus,   dy/dx = d/dx [sec(2x)] = sec(2x)tan(2x) · 2 = 2 sec(2x)tan(2x). Step 2. Find the second derivative (d²y/dx²). Differentiate dy/dx = 2 sec(2x)tan(2x) with respect to x. Use the product rule:   d/dx [uv] = u'v + uv' Let   u(x) = sec(2x)  and  v(x) = tan(2x). First, compute u'(x):   u'(x) = d/dx [sec(2x)] = 2 sec(2x)tan(2x). Next, compute v'(x):   v'(x) = d/dx [tan(2x)] = 2 sec²(2x). Now, applying the product rule:   d/dx [sec(2x)·tan(2x)] = u'·v + u·v'                 = [2 sec(2x)tan(2x)]·tan(2x) + sec(2x)·[2 sec²(2x)]                 = 2 sec(2x)tan²(2x) + 2 sec(2x)sec²(2x)                 = 2 sec(2x)[tan²(2x) + sec²(2x)]. Now remember the factor 2 from the original derivative:   d²y/dx² = 2 × {2 sec(2x)[tan²(2x) + sec²(2x)]}        = 4 sec(2x)[tan²(2x) + sec²(2x)]. Step 3. Optional Simplification We can simplify further by recalling the trigonometric identity:   tan²(θ) = sec²(θ) − 1. In our case (θ = 2x):   tan²(2x) = sec²(2x) − 1. Substitute this into the expression:   tan²(2x) + sec²(2x) = (sec²(2x) − 1) + sec²(2x)                  = 2 sec²(2x) − 1. Thus, the second derivative becomes:   d²y/dx² = 4 sec(2x)[2 sec²(2x) − 1]. Alternatively, we can expand this as:   d²y/dx² = 8 sec³(2x) − 4 sec(2x). Final Answer The second derivative of y = sec(2x) is   d²y/dx² = 4 sec(2x)[tan²(2x) + sec²(2x)]       = 4 sec(2x)[2 sec²(2x) − 1]  or  8 sec³(2x) − 4 sec(2x).