Let's solve each part step by step: --- ### **(2.1) If \( x = b\sin{3t} \) and \( y = b\cos{3t} \), find \(\frac{d^2y}{dx^2}\):** #### **Step 1: Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)** \[ x = b\sin{3t} \implies \frac{dx}{dt} = b \cdot 3\cos{3t} = 3b\cos{3t} \] \[ y = b\cos{3t} \implies \frac{dy}{dt} = b \cdot (-3\sin{3t}) = -3b\sin{3t} \] #### **Step 2: Find \(\frac{dy}{dx}\) using the chain rule** \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-3b\sin{3t}}{3b\cos{3t}} = -\tan{3t} \] #### **Step 3: Differentiate \(\frac{dy}{dx}\) with respect to \(x\) to get \(\frac{d^2y}{dx^2}\)** First, find \(\frac{d}{dt}(-\tan{3t})\): \[ \frac{d}{dt}(-\tan{3t}) = -3\sec^2{3t} \] Now, \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}(-\tan{3t})}{\frac{dx}{dt}} = \frac{-3\sec^2{3t}}{3b\cos{3t}} \] \[ \sec^2{3t} = \frac{1}{\cos^2{3t}} \] So, \[ \frac{d^2y}{dx^2} = \frac{-3}{3b\cos{3t}}\cdot\frac{1}{\cos^2{3t}} = \frac{-1}{b\cos^3{3t}} \] --- ### **(2.2) If \( x^4 + y^4 = 16 \), find \(\frac{d^2y}{dx^2}\):** #### **Step 1: Differentiate both sides w.r.t \(x\):** \[ \frac{d}{dx}(x^4 + y^4) = \frac{d}{dx}(16) \] \[ 4x^3 + 4y^3\frac{dy}{dx} = 0 \] \[ 4y^3\frac{dy}{dx} = -4x^3 \] \[ \frac{dy}{dx} = -\frac{x^3}{y^3} \] #### **Step 2: Differentiate again w.r.t \(x\):** Apply the quotient rule and chain rule: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x^3}{y^3}\right) \] Let’s use the quotient rule: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \] where \(u = -x^3\), \(v = y^3\) - \(\frac{du}{dx} = -3x^2\) - \(\frac{dv}{dx} = 3y^2\frac{dy}{dx}\) So, \[ \frac{d^2y}{dx^2} = \frac{y^3(-3x^2) - (-x^3)\cdot 3y^2\frac{dy}{dx}}{y^6} \] \[ = \frac{-3x^2y^3 + 3x^3y^2\frac{dy}{dx}}{y^6} \] \[ = \frac{-3x^2y^3}{y^6} + \frac{3x^3y^2}{y^6}\frac{dy}{dx} \] \[ = -\frac{3x^2}{y^3} + \frac{3x^3}{y^4}\frac{dy}{dx} \] Recall from above: \(\frac{dy}{dx} = -\frac{x^3}{y^3}\): \[ = -\frac{3x^2}{y^3} + \frac{3x^3}{y^4} \left(-\frac{x^3}{y^3}\right) \] \[ = -\frac{3x^2}{y^3} - \frac{3x^6}{y^7} \] --- ### **(2.3) \( y = \frac{1}{(1 + \tan{x})^2} \), find \(\frac{d^2y}{dx^2}\):** Let’s set \(u = 1 + \tan{x}\), so \(y = u^{-2}\). #### **Step 1: First Derivative** \[ \frac{dy}{dx} = -2u^{-3}\cdot \sec^2{x} = -2\frac{\sec^2{x}}{(1 + \tan{x})^3} \] #### **Step 2: Second Derivative** Use the product and chain rules: Let’s use the quotient rule: \[ \frac{d}{dx}\left(\frac{\sec^2{x}}{(1+\tan{x})^3}\right) \] Let \(f(x) = \sec^2{x}\), \(g(x) = (1+\tan{x})^3\) \[ \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{g\frac{df}{dx} - f\frac{dg}{dx}}{g^2} \] - \(\frac{df}{dx} = 2\sec^2{x}\tan{x}\) - \(\frac{dg}{dx} = 3(1+\tan{x})^2 \cdot \sec^2{x}\) So, \[ \frac{d^2y}{dx^2} = -2 \frac{(1+\tan{x})^3[2\sec^2{x}\tan{x}] - \sec^2{x}[3(1+\tan{x})^2\sec^2{x}]}{(1+\tan{x})^6} \] Let’s expand and simplify: \[ = -2 \frac{2(1+\tan{x})^3\sec^2{x}\tan{x} - 3(1+\tan{x})^2\sec^4{x}}{(1+\tan{x})^6} \] \[ = -2 \frac{(1+\tan{x})^2\sec^2{x}[2(1+\tan{x})\tan{x} - 3\sec^2{x}]}{(1+\tan{x})^6} \] \[ = -2 \frac{\sec^2{x}}{(1+\tan{x})^4}[2(1+\tan{x})\tan{x} - 3\sec^2{x}] \] --- ## **Final Answers:** ### **(2.1)** \[ \boxed{\frac{d^2y}{dx^2} = -\frac{1}{b\cos^3{3t}}} \] --- ### **(2.2)** \[ \boxed{\frac{d^2y}{dx^2} = -\frac{3x^2}{y^3} - \frac{3x^6}{y^7}} \] --- ### **(2.3)** \[ \boxed{ \frac{d^2y}{dx^2} = -2 \frac{\sec^2{x}}{(1+\tan{x})^4}\left[2(1+\tan{x})\tan{x} - 3\sec^2{x}\right] } \] Let me know if you need any step explained further!