1. Problem: Differentiate the given functions with respect to their variables. (i) $y = \sqrt[3]{3x^{2} + 1}$ 1. Rewrite as $y = (3x^{2}+1)^{\frac{1}{3}}$. 2. Use chain rule: $\frac{dy}{dx} = \frac{1}{3}(3x^{2}+1)^{-\frac{2}{3}} \cdot 6x = \frac{2x}{(3x^{2}+1)^{\frac{2}{3}}}$. (ii) $y = \frac{1}{\sqrt[3]{4 - 3x}} = (4 - 3x)^{-\frac{1}{3}}$ 1. Differentiate: $\frac{dy}{dx} = -\frac{1}{3}(4 - 3x)^{-\frac{4}{3}} \cdot (-3) = \frac{1}{(4 - 3x)^{\frac{4}{3}}}$. (iii) $y = \sqrt{ax^{2} + bx + c} = (ax^{2} + bx + c)^{\frac{1}{2}}$ 1. Differentiate: $\frac{dy}{dx} = \frac{1}{2}(ax^{2} + bx + c)^{-\frac{1}{2}} \cdot (2ax + b) = \frac{2ax + b}{2\sqrt{ax^{2} + bx + c}}$. (iv) $y = \sin\left(\frac{\theta}{2}\right) + \cos 5\theta$ 1. Derivative w.r.t $\theta$: $\frac{dy}{d\theta} = \cos\left(\frac{\theta}{2}\right) \cdot \frac{1}{2} - 5\sin 5\theta = \frac{1}{2} \cos\left(\frac{\theta}{2}\right) - 5\sin 5\theta$. (v) $y = \cos^{4} \theta = (\cos \theta)^{4}$ 1. Derivative: $\frac{dy}{d\theta} = 4(\cos \theta)^{3} (-\sin \theta) = -4 \cos^{3} \theta \sin \theta$. (vi) $y = \sec \sqrt{x} = \sec(x^{\frac{1}{2}})$ 1. Derivative: $\frac{dy}{dx} = \sec \left(\sqrt{x} \right) \tan \left(\sqrt{x} \right) \cdot \frac{1}{2\sqrt{x}} = \frac{\sec(\sqrt{x}) \tan(\sqrt{x})}{2\sqrt{x}}$. (vii) $y = \sqrt{\tan \theta} = (\tan \theta)^{\frac{1}{2}}$ 1. Derivative: $\frac{dy}{d\theta} = \frac{1}{2} (\tan \theta)^{ -\frac{1}{2}} \cdot \sec^{2} \theta = \frac{\sec^{2} \theta}{2 \sqrt{\tan \theta}}$. (viii) $y = \sqrt{\tan e^{x^{2}}} = (\tan e^{x^{2}})^{\frac{1}{2}}$ 1. Derivative using chain rule: $$\frac{dy}{dx} = \frac{1}{2}(\tan e^{x^{2}})^{-\frac{1}{2}} \cdot \sec^{2} e^{x^{2}} \cdot e^{x^{2}} \cdot 2x = \frac{\sec^{2} e^{x^{2}} \cdot e^{x^{2}} \cdot x}{\sqrt{\tan e^{x^{2}}}}$$. (ix) $y = \cos x^{\circ}$ Treat $x^{\circ}$ as degrees; conversion: $x^{\circ} = \frac{\pi x}{180}$ radians. 1. Derivative: $$\frac{dy}{dx} = - \sin \left( \frac{\pi x}{180} \right) \cdot \frac{\pi}{180}$$. (x) $y = \cos^{2} x^{2} = (\cos x^{2})^{2}$ 1. Derivative: $$\frac{dy}{dx} = 2 \cos x^{2} \cdot (-\sin x^{2}) \cdot 2x = -4x \cos x^{2} \sin x^{2}$$. (xi) $y = \sin^{3} x^{3} = (\sin x^{3})^{3}$ 1. Derivative: $$\frac{dy}{dx} = 3 (\sin x^{3})^{2} \cdot \cos x^{3} \cdot 3x^{2} = 9x^{2} \sin^{2} x^{3} \cos x^{3}$$. --- 2. Problem: Differentiate the given exponential and logarithmic functions. (i) $y = e^{\sqrt{x}} = e^{x^{1/2}}$ 1. Derivative: $$\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2 \sqrt{x}}$$. (ii) $y = \sqrt{\frac{1}{e^{x}}} = e^{-x/2}$ 1. Derivative: $$\frac{dy}{dx} = -\frac{1}{2} e^{-x/2}$$. (iii) $y = \sqrt{e^{\sqrt{x}}} = e^{\frac{\sqrt{x}}{2}}$ 1. Derivative: $$\frac{dy}{dx} = e^{\frac{\sqrt{x}}{2}} \cdot \frac{1}{4 \sqrt{x}} = \frac{e^{\frac{\sqrt{x}}{2}}}{4 \sqrt{x}}$$. (v) $y = e^{\sin^{2} x}$ 1. Derivative: $$\frac{dy}{dx} = e^{\sin^{2} x} \cdot 2 \sin x \cos x = 2 e^{\sin^{2} x} \sin x \cos x$$. (vi) $y = e^{\sin (\cos^{-1} 2x)}$ 1. Let $u = \sin(\cos^{-1} 2x)$, use identity $$\sin(\cos^{-1} t) = \sqrt{1 - t^{2}}$$ for $t=2x$. 2. So $u = \sqrt{1 - 4x^{2}}$. 3. Then $y = e^{u}$ and $$\frac{dy}{dx} = e^{u} \cdot \frac{du}{dx} = e^{\sqrt{1 - 4x^{2}}} \cdot \frac{-4x}{\sqrt{1 - 4x^{2}}} = -\frac{4x e^{\sqrt{1 - 4x^{2}}}}{\sqrt{1 - 4x^{2}}}$$. (vii) $y = e^{x} \log 2x \cdot e^{2x} = e^{3x} \log 2x$ 1. Using product rule: $$\frac{dy}{dx} = e^{3x} \log 2x \cdot 3 + e^{3x} \cdot \frac{1}{2x \ln 10} \cdot 2 = 3 e^{3x} \log 2x + \frac{e^{3x}}{x \ln 10}$$. --- 3. Problem: Differentiate logarithmic and related functions. (i) $y = \ln(\cos 2\theta)$ $$\frac{dy}{d\theta} = \frac{-2\sin 2\theta}{\cos 2\theta} = -2 \tan 2\theta$$. (ii) $y = \ln (\sin 2x)$ $$\frac{dy}{dx} = \frac{2 \cos 2x}{\sin 2x} = 2 \cot 2x$$. (iii) $y = \ln v \sqrt{x} = \ln (v \sqrt{x}) = \ln v + \frac{1}{2} \ln x$ $$\frac{dy}{dx} = 0 + \frac{1}{2x} = \frac{1}{2x}$$ assuming $v$ constant. (v) $y = \log_{10} x = \frac{\ln x}{\ln 10}$ $$\frac{dy}{dx} = \frac{1}{x \ln 10}$$. (vi) $y = \log_{10} 3x = \log_{10} 3 + \log_{10} x$ $$\frac{dy}{dx} = 0 + \frac{1}{x \ln 10} = \frac{1}{x \ln 10}$$. (vii) $y = \ln (ax^{2} + bx + c)$ $$\frac{dy}{dx} = \frac{2ax + b}{ax^{2} + bx + c}$$. (viii) $y = 10^{\ln (\sin x)}$ 1. Rewrite $10^{\ln (\sin x)} = e^{\ln (\sin x) \ln 10}$. 2. Derivative: $$\frac{dy}{dx} = 10^{\ln (\sin x)} \cdot \ln 10 \cdot \frac{\cos x}{\sin x} = 10^{\ln (\sin x)} \ln 10 \cot x$$. --- 4. Problem: Differentiate complex expressions involving exponentials and logarithms. (i) $y = e^{2 \ln (\tan 5x)} = e^{\ln (\tan^{2} 5x)} = \tan^{2} 5x$ $$\frac{dy}{dx} = 2 \tan 5x \cdot \sec^{2} 5x \cdot 5 = 10 \tan 5x \sec^{2} 5x$$. (ii) $y = e^{\frac{5}{\ln (\tan 5x)}}$ $$\frac{dy}{dx} = y \cdot \frac{-5}{(\ln \tan 5x)^{2}} \cdot \frac{1}{\tan 5x} \cdot \sec^{2} 5x \cdot 5 = -\frac{25 y \sec^{2} 5x}{\tan 5x (\ln \tan 5x)^{2}}$$. (iii) $y = \cos(\ln x) + \ln (\tan x)$ $$\frac{dy}{dx} = -\sin(\ln x) \cdot \frac{1}{x} + \frac{1}{\tan x} \cdot \sec^{2} x = -\frac{\sin(\ln x)}{x} + \frac{\sec^{2} x}{\tan x}$$. (iv) $y = (1 + \sin 2x)^{2}$ $$\frac{dy}{dx} = 2 (1 + \sin 2x) \cdot 2 \cos 2x = 4 (1 + \sin 2x) \cos 2x$$. --- 5. Problem: Differentiate Type-II functions. (i) $y = \sin^{2} \{\ln (\sec x)\}$ $$\frac{dy}{dx} = 2 \sin \ln(\sec x) \cdot \cos \ln(\sec x) \cdot \frac{d}{dx} \ln(\sec x) = \sin(2 \ln(\sec x)) \cdot \sec x \tan x / \sec x = \sin(2 \ln(\sec x)) \tan x$$. (ii) $y = \sin^{2} (\ln \cos x)$ $$\frac{dy}{dx} = \sin(2 \ln \cos x) \cdot \frac{-\tan x}{\cos x} \Rightarrow \sin(2 \ln \cos x) (-\tan x)$$ (simplified). (iii) $y = \ln (\sin x^{2}) = \ln (\sin (x^{2}))$ $$\frac{dy}{dx} = \frac{\cos x^{2} \cdot 2x}{\sin x^{2}} = 2x \cot x^{2}$$. (iv) $y = (\ln (\sin x^{2}))^{n}$ $$\frac{dy}{dx} = n (\ln (\sin x^{2}))^{n-1} \cdot \frac{2x \cot x^{2}}{1} = 2n x (\ln (\sin x^{2}))^{n-1} \cot x^{2}$$. (v) $y = \sin^{2} \{\ln (x^{2})\} = \sin^{2} (2 \ln x)$ $$\frac{dy}{dx} = \sin(4 \ln x) \cdot \frac{2}{x} = \frac{2}{x} \sin(4 \ln x)$$. (vi) $y = \sin \{\ln (\sec x)\}$ $$\frac{dy}{dx} = \cos \ln(\sec x) \cdot \sec x \tan x / \sec x = \cos (\ln (\sec x)) \tan x$$. --- 6. Problem: Differentiate Type-III functions. (i) $y = x^{\circ} \cos x^{\circ}$, i.e. $x$ in degrees. 1. Use $x^{\circ} = \frac{\pi x}{180}$ radians. 2. Derivative: $$\frac{dy}{dx} = \cos \frac{\pi x}{180} + x \left(- \sin \frac{\pi x}{180} \cdot \frac{\pi}{180} \right) = \cos \frac{\pi x}{180} - \frac{\pi x}{180} \sin \frac{\pi x}{180}$$. (ii) $y = \theta^{\circ} \sin \theta^{\circ}$ similar conversion. (iii) $y = x \sqrt{x^{2} + a^{2}}$ $$\frac{dy}{dx} = \sqrt{x^{2} + a^{2}} + x \cdot \frac{x}{\sqrt{x^{2}+a^{2}}} = \frac{2x^{2} + a^{2}}{\sqrt{x^{2}+a^{2}}}$$. (iv) $y = x \sqrt{\sin x}$ $$\frac{dy}{dx} = \sqrt{\sin x} + x \cdot \frac{1}{2 \sqrt{\sin x}} \cos x = \sqrt{\sin x} + \frac{x \cos x}{2 \sqrt{\sin x}}$$. (v) $y = x^{n} \ln 2x = x^{n} (\ln 2 + \ln x)$ $$\frac{dy}{dx} = n x^{n-1} \ln 2x + x^{n} \cdot \frac{1}{x} = n x^{n-1} (\ln 2 + \ln x) + x^{n-1} = x^{n-1} (n \ln 2x + 1)$$. (vi) $y = \sin (e^{\sqrt{1-x}})$ $$\frac{dy}{dx} = \cos (e^{\sqrt{1-x}}) \cdot e^{\sqrt{1-x}} \cdot \frac{d}{dx} \sqrt{1-x} = \cos (e^{\sqrt{1-x}}) e^{\sqrt{1-x}} \cdot \frac{-1}{2 \sqrt{1-x}}$$. (vii) $y = e^{ax} \tan^{2} x$ $$\frac{dy}{dx} = a e^{ax} \tan^{2} x + e^{ax} \cdot 2 \tan x \sec^{2} x = e^{ax} [ a \tan^{2} x + 2 \tan x \sec^{2} x ]$$. (viii) $y = 2^{x} \ln \left( \frac{1}{1-x} \right) = 2^{x} \ln (1 - x)^{-1} = -2^{x} \ln (1-x)$ $$\frac{dy}{dx} = -2^{x} \ln (1-x) \ln 2 + 2^{x} \cdot \frac{1}{1-x} = 2^{x} \left( \frac{1}{1-x} - \ln (1-x) \ln 2 \right)$$. Final answers include detailed differentiation of each function with all steps.