1. **Problem 4.1:** Find the derivative of the function $f(x) = (2x - 1)^2 \sin(2x)$. 2. **Formula and rules:** Use the product rule for derivatives: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ where $u(x) = (2x - 1)^2$ and $v(x) = \sin(2x)$. Also use the chain rule for derivatives of composite functions. 3. **Step-by-step solution:** - Derivative of $u(x)$: $$u'(x) = 2(2x - 1) \cdot \frac{d}{dx}(2x - 1) = 2(2x - 1) \cdot 2 = 4(2x - 1)$$ - Derivative of $v(x)$: $$v'(x) = \cos(2x) \cdot \frac{d}{dx}(2x) = 2\cos(2x)$$ - Apply product rule: $$f'(x) = u'(x)v(x) + u(x)v'(x) = 4(2x - 1) \sin(2x) + (2x - 1)^2 \cdot 2\cos(2x)$$ 4. **Final answer for 4.1:** $$f'(x) = 4(2x - 1) \sin(2x) + 2(2x - 1)^2 \cos(2x)$$ --- 1. **Problem 4.2:** Find the derivative of the function $$f(x) = x^2 + \frac{2}{\sqrt{x^2}}$$ 2. **Rewrite the function:** $$f(x) = x^2 + 2x^{-1}$$ since $\sqrt{x^2} = |x|$ but for derivative purposes and $x \neq 0$, treat as $x$. 3. **Derivative rules:** Use power rule: $$\frac{d}{dx} x^n = nx^{n-1}$$ 4. **Step-by-step solution:** - Derivative of $x^2$: $$2x$$ - Derivative of $2x^{-1}$: $$2 \cdot (-1) x^{-2} = -2x^{-2}$$ - Combine: $$f'(x) = 2x - 2x^{-2} = 2x - \frac{2}{x^2}$$ 5. **Final answer for 4.2:** $$f'(x) = 2x - \frac{2}{x^2}$$ --- 1. **Problem 4.3:** Find the equation of the tangent line to the parabola $$y = x^2 - 4$$ at the point where $x = 1$. 2. **Formula and rules:** The slope of the tangent line is the derivative evaluated at $x=1$. The tangent line equation is $$y - y_1 = m(x - x_1)$$ where $m = f'(x_1)$ and $(x_1, y_1)$ is the point on the curve. 3. **Step-by-step solution:** - Derivative of $y = x^2 - 4$: $$y' = 2x$$ - Evaluate slope at $x=1$: $$m = 2(1) = 2$$ - Find point on curve at $x=1$: $$y_1 = 1^2 - 4 = -3$$ - Equation of tangent line: $$y - (-3) = 2(x - 1)$$ $$y + 3 = 2x - 2$$ $$y = 2x - 5$$ 4. **Final answer for 4.3:** $$y = 2x - 5$$