1. The problem asks us to find \(\frac{dy}{dx}\) given that \(\tan(x y^{2}) = (2x + y)^{3}\). We must differentiate both sides implicitly with respect to \(x\). 2. To differentiate \(f(x) = 4^{x^{2} \ln(x^{2} + \sqrt[3]{x})}\), we use the chain rule and logarithmic differentiation. 3. Find the derivative of \(f(x) = \cot^{-1} \left( \frac{1}{1 + (\tan^{-1} x)^{2}} \right)\) by applying the chain rule and derivatives of inverse trigonometric functions. 4. Find the equation of the tangent line to \(y = \sin^{-1} \left( \frac{x}{3} \right)\) at the origin (0,0). 5. Find the equation of the normal line to \(y = (\log(x-1))^{\cos \sqrt[3]{x-1}}\) at \(x = \sqrt{\pi}\). 6. Evaluate the limit \(\lim_{x \to +\infty} \frac{\ln (x + e^{x})}{3x}\). 7. Evaluate the limit \(\lim_{x \to 0} \frac{\tan^{-1}(2x)}{\sin^{-1}(3x)}\). --- 1. Differentiate implicitly: Start with \(\tan(x y^{2}) = (2x + y)^{3}\). Apply derivative on both sides: \[ \sec^{2}(x y^{2}) \cdot \frac{d}{dx}(x y^{2}) = 3(2x + y)^{2} \cdot \frac{d}{dx}(2x + y) \] Calculate each derivative: \[ \frac{d}{dx}(x y^{2}) = y^{2} + x \cdot 2y \frac{dy}{dx} \] \[ \frac{d}{dx}(2x + y) = 2 + \frac{dy}{dx} \] So: \[ \sec^{2}(x y^{2}) (y^{2} + 2x y \frac{dy}{dx}) = 3 (2x + y)^{2} (2 + \frac{dy}{dx}) \] This is the implicit derivative, not simplified. 2. For \(f(x) = 4^{x^{2} \ln(x^{2} + \sqrt[3]{x})}\), set \(u = x^{2} \ln(x^{2} + x^{1/3})\). Then: \[ f(x) = 4^{u} \implies f'(x) = 4^{u} \ln(4) \cdot u' \] Find derivative of \(u\): \[ u = x^{2} \ln(x^{2} + x^{1/3}) \] Using product rule: \[ u' = 2x \ln(x^{2} + x^{1/3}) + x^{2} \cdot \frac{1}{x^{2} + x^{1/3}} \cdot (2x + \frac{1}{3} x^{-2/3}) \] 3. For \(f(x) = \cot^{-1} \left( \frac{1}{1 + (\tan^{-1} x)^{2}} \right)\), set: \(v = \frac{1}{1 + (\tan^{-1} x)^{2}}\). Derivative: \[ f'(x) = - \frac{1}{1 + v^{2}} \cdot v' \] Now, \[ v' = - \frac{2 (\tan^{-1} x) \cdot \frac{1}{1 + x^{2}}}{(1 + (\tan^{-1} x)^{2})^{2}} \] Combine these for final derivative (do not simplify). 4. For tangent line to \(y = \sin^{-1} \left( \frac{x}{3} \right)\) at \(x=0\): First, find \(y(0) = \sin^{-1}(0) = 0\). Find \(y'\): \[ y' = \frac{1}{\sqrt{1 - (x/3)^{2}}} \cdot \frac{1}{3} \] At \(x=0\): \[ y'(0) = \frac{1}{3} \] Tangent line equation: \[ y - 0 = \frac{1}{3} (x - 0) \implies y = \frac{x}{3} \] 5. For normal line to \(y = (\log(x-1))^{\cos \sqrt[3]{x-1}}\) at \(x = \sqrt{\pi}\): Find \(y\) at \(x=\sqrt{\pi}\): \[ y = (\log(\sqrt{\pi} -1))^{\cos ( (\sqrt{\pi} -1)^{1/3} )} \] Find \(dy/dx\) (use logarithmic differentiation and chain rule). Then normal line slope is negative reciprocal: \[ m_{normal} = - \frac{1}{y'(\sqrt{\pi})} \] Equation of normal line: \[ y - y(\sqrt{\pi}) = m_{normal} (x - \sqrt{\pi}) \] 6. Evaluate \(\lim_{x \to +\infty} \frac{\ln (x + e^{x})}{3x}\). Note \(e^{x}\) dominates \(x\) for large \(x\). So \(\ln(x + e^{x}) \sim \ln(e^{x}) = x\). Therefore, \[ \lim_{x \to +\infty} \frac{\ln (x + e^{x})}{3x} = \lim_{x \to +\infty} \frac{x}{3x} = \frac{1}{3} \] 7. Evaluate \(\lim_{x \to 0} \frac{\tan^{-1}(2x)}{\sin^{-1}(3x)}\). Using small angle approximations: \[ \tan^{-1}(2x) \approx 2x, \quad \sin^{-1}(3x) \approx 3x \] So limit is: \[ \lim_{x \to 0} \frac{2x}{3x} = \frac{2}{3} \] --- Final answers: 1. \( \frac{dy}{dx} \) implicit relation: \[ \sec^{2}(x y^{2}) (y^{2} + 2 x y y') = 3 (2 x + y)^{2} (2 + y') \] 2. \( f'(x) = 4^{x^{2} \ln(x^{2} + x^{1/3})} \ln 4 \left[ 2 x \ln(x^{2} + x^{1/3}) + \frac{x^{2} (2 x + \frac{1}{3} x^{-2/3})}{x^{2} + x^{1/3}} \right] \) 3. \( f'(x) = - \frac{1}{1 + \left( \frac{1}{1 + (\tan^{-1} x)^{2}} \right)^{2}} \times \left[ - \frac{2 (\tan^{-1} x) \cdot \frac{1}{1 + x^{2}}}{(1 + (\tan^{-1} x)^{2})^{2}} \right] \) 4. Tangent line: \( y = \frac{x}{3} \) 5. Normal line: \[ y - (\log(\sqrt{\pi} -1))^{\cos ((\sqrt{\pi} -1)^{1/3})} = -\frac{1}{y'(\sqrt{\pi})} (x - \sqrt{\pi}) \] 6. \( \lim_{x \to +\infty} \frac{\ln (x + e^{x})}{3x} = \frac{1}{3} \) 7. \( \lim_{x \to 0} \frac{\tan^{-1}(2x)}{\sin^{-1}(3x)} = \frac{2}{3} \)