1. Find the derivative of the function $f(x) = (2x - 1)^2 \sin(2x)$.\n\nStep 1: State the problem: Differentiate $f(x) = (2x - 1)^2 \sin(2x)$.\nStep 2: Use the product rule: If $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.\nStep 3: Let $u = (2x - 1)^2$ and $v = \sin(2x)$.\nStep 4: Compute $u' = 2(2x - 1) \cdot 2 = 4(2x - 1)$ using the chain rule.\nStep 5: Compute $v' = \cos(2x) \cdot 2 = 2\cos(2x)$.\nStep 6: Apply product rule:\n$$f'(x) = 4(2x - 1) \sin(2x) + (2x - 1)^2 \cdot 2 \cos(2x)$$\n\n2. Find the derivative of the function $f(x) = \frac{x^2 - x + 2}{\sqrt{x^2}}$.\n\nStep 1: State the problem: Differentiate $f(x) = \frac{x^2 - x + 2}{\sqrt{x^2}}$.\nStep 2: Simplify denominator: $\sqrt{x^2} = |x|$. For $x > 0$, $\sqrt{x^2} = x$.\nStep 3: So for $x > 0$, $f(x) = \frac{x^2 - x + 2}{x} = x - 1 + \frac{2}{x}$.\nStep 4: Differentiate term by term:\n$$f'(x) = 1 - 0 - \frac{2}{x^2} = 1 - \frac{2}{x^2}$$\n\n3. Find the equation of the tangent line to the parabola $y = x^2 - 4$ at $x = 1$.\n\nStep 1: State the problem: Find tangent line to $y = x^2 - 4$ at $x=1$.\nStep 2: Find derivative $y' = 2x$.\nStep 3: Evaluate slope at $x=1$: $m = 2(1) = 2$.\nStep 4: Find point on curve: $y = 1^2 - 4 = -3$. Point is $(1, -3)$.\nStep 5: Use point-slope form: $y - y_1 = m(x - x_1)$\n$$y + 3 = 2(x - 1)$$\n$$y = 2x - 2 - 3 = 2x - 5$$\n\nFinal answers:\n1. $$f'(x) = 4(2x - 1) \sin(2x) + 2(2x - 1)^2 \cos(2x)$$\n2. $$f'(x) = 1 - \frac{2}{x^2} \quad \text{for } x > 0$$\n3. Tangent line equation: $$y = 2x - 5$$