Let's solve each question step by step: --- ### **Question 1** **Given:** \( S = 3t^3 - 2t^2 + 4t - 1 \) #### **1.1 Velocity at \( t = 0 \) and \( t = 1.5 \) s** Velocity is the first derivative of displacement with respect to time: \[ v = \frac{dS}{dt} = \frac{d}{dt}(3t^3 - 2t^2 + 4t - 1) = 9t^2 - 4t + 4 \] - **At \( t = 0 \):** \[ v = 9(0)^2 - 4(0) + 4 = 4 \text{ m/s} \] - **At \( t = 1.5 \):** \[ v = 9(1.5)^2 - 4(1.5) + 4 = 9(2.25) - 6 + 4 = 20.25 - 6 + 4 = 18.25 \text{ m/s} \] #### **1.2 Acceleration at \( t = 0 \) and \( t = 1.5 \) s** Acceleration is the derivative of velocity: \[ a = \frac{dv}{dt} = \frac{d}{dt}(9t^2 - 4t + 4) = 18t - 4 \] - **At \( t = 0 \):** \[ a = 18(0) - 4 = -4 \text{ m/s}^2 \] - **At \( t = 1.5 \):** \[ a = 18(1.5) - 4 = 27 - 4 = 23 \text{ m/s}^2 \] --- ### **Question 2** **Given:** \( f(x) = x^2 + 2x - 1 \) #### **2.1 Equation of the tangent at \( x = 1 \):** - Find \( f'(x) \): \[ f'(x) = 2x + 2 \] - At \( x = 1 \): \( f'(1) = 2(1) + 2 = 4 \) - \( f(1) = (1)^2 + 2(1) - 1 = 2 \) - Equation of tangent: \( y - f(1) = f'(1)(x - 1) \) \[ y - 2 = 4(x - 1) \implies y = 4x - 2 \] #### **2.2 Equation of normal at \( x = 1 \):** Slope of normal = \( -\frac{1}{4} \) - Equation: \( y - 2 = -\frac{1}{4}(x - 1) \) \[ y - 2 = -\frac{1}{4}(x - 1) \implies y = -\frac{1}{4}x + \frac{1}{4} + 2 \implies y = -\frac{1}{4}x + \frac{9}{4} \] --- ### **Question 3** **Given:** \( f(x) = \frac{x}{\sqrt{x^2+1}} \) #### **3.1 Equation of tangent at \( x = 1 \):** - \( f(1) = \frac{1}{\sqrt{2}} \) - \( f'(x) \) using quotient rule: \[ f'(x) = \frac{\sqrt{x^2+1}(1) - x \cdot \frac{x}{\sqrt{x^2+1}}}{x^2+1} = \frac{\sqrt{x^2+1} - \frac{x^2}{\sqrt{x^2+1}}}{x^2+1} = \frac{x^2+1 - x^2}{(x^2+1)^{3/2}} = \frac{1}{(x^2+1)^{3/2}} \] - At \( x = 1 \): \( f'(1) = \frac{1}{(2)^{3/2}} = \frac{1}{2\sqrt{2}} \) - Tangent: \( y - \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}} (x - 1) \) #### **3.2 Equation of normal at \( x = 1 \):** Slope of normal = \( -2\sqrt{2} \) Normal: \( y - \frac{1}{\sqrt{2}} = -2\sqrt{2}(x - 1) \) --- ### **Question 4** #### **4.1 \( y = x\sin x \)** Product rule: \[ y' = x\cos x + \sin x \] #### **4.2 \( y = \frac{\cos(2x)}{\sin(4x)} \)** Quotient rule: \[ y' = \frac{\sin(4x)(-\sin(2x) \cdot 2) - \cos(2x) \cdot \cos(4x) \cdot 4}{\sin^2(4x)} \] \[ y' = \frac{-2\sin(4x)\sin(2x) - 4\cos(2x)\cos(4x)}{\sin^2(4x)} \] --- ### **Question 5** #### **5.1 \( y = \frac{(\ln x)\cos 3x}{x^2\sin 3x} \)** Let \( y = \frac{u}{v} \), where \( u = (\ln x)\cos 3x \), \( v = x^2\sin 3x \): \[ y' = \frac{v u' - u v'}{v^2} \] - \( u' = \frac{1}{x}\cos 3x + (\ln x)(-3\sin 3x) \) - \( v' = 2x\sin 3x + x^2(3\cos 3x) \) Plug and simplify as needed. #### **5.2 \( y = \frac{e^{4x}\sin(2\theta)}{5} \)** \[ y' = \frac{e^{4x}\sin(2\theta)}{5} \cdot 4 = \frac{4e^{4x}\sin(2\theta)}{5} \] --- ### **Question 6** #### **6.1 \( f(x) = \tan^{-1}(\sqrt{x}) \)** Let \( u = \sqrt{x} \), \( f(x) = \tan^{-1}(u) \): \[ f'(x) = \frac{1}{1+u^2} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}(1+x)} \] --- ## **Summary of Final Answers:** ### **Q1.** 1.1. At \( t=0 \), \( v = 4 \) m/s; at \( t=1.5 \), \( v = 18.25 \) m/s 1.2. At \( t=0 \), \( a = -4 \) m/s\(^2\); at \( t=1.5 \), \( a = 23 \) m/s\(^2\) ### **Q2.** 2.1. Tangent: \( y = 4x - 2 \) 2.2. Normal: \( y = -\frac{1}{4}x + \frac{9}{4} \) ### **Q3.** 3.1. Tangent: \( y - \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}} (x - 1) \) 3.2. Normal: \( y - \frac{1}{\sqrt{2}} = -2\sqrt{2}(x - 1) \) ### **Q4.** 4.1. \( y' = x\cos x + \sin x \) 4.2. \( y' = \frac{-2\sin(4x)\sin(2x) - 4\cos(2x)\cos(4x)}{\sin^2(4x)} \) ### **Q5.** 5.1. Use quotient and product rule as shown above. 5.2. \( y' = \frac{4e^{4x}\sin(2\theta)}{5} \) ### **Q6.** 6.1. \( f'(x) = \frac{1}{2\sqrt{x}(1 + x)} \) --- Let me know if you want detailed simplification for any specific part!

Certainly! Here is a concise 6-line pure theory on derivatives and their applications: 1. **Derivative Definition:** The derivative of a function measures the rate at which the function's value changes with respect to its independent variable, representing the slope of the tangent line at any point. 2. **Basic Rules:** Key derivative rules include the power rule, product rule, quotient rule, and chain rule, which facilitate differentiation of algebraic and composite functions. 3. **Velocity and Acceleration:** In physics, the derivative of displacement with respect to time gives velocity, while the derivative of velocity gives acceleration, describing motion dynamics. 4. **Tangents and Normals:** The slope of the tangent line to a curve at a point is given by the derivative at that point; the normal is perpendicular to the tangent, with its slope being the negative reciprocal. 5. **Inverse Trigonometric Derivatives:** The derivatives of inverse functions like \( \tan^{-1} x \), \( \sin^{-1} x \), etc., are essential in solving integrals and solving equations involving inverse trig functions. 6. **Logarithmic Differentiation:** A technique used to differentiate functions where variables appear as exponents or products inside logarithms, simplifying complex differentiation tasks.