1. Differentiate $x^2 \cot x$ using the product rule: $$\frac{d}{dx}(x^2 \cot x) = 2x \cot x + x^2(-\csc^2 x) = 2x \cot x - x^2 \csc^2 x$$ 2. Differentiate $\frac{\csc x}{x^3}$ using the quotient rule: Let $u=\csc x, v=x^3$ then $$u' = -\csc x \cot x, v' = 3x^2$$ $$\frac{d}{dx}\left(\frac{\csc x}{x^3}\right) = \frac{v u' - u v'}{v^2} = \frac{x^3(-\csc x \cot x) - \csc x (3x^2)}{x^6} = \frac{-x^3 \csc x \cot x - 3x^2 \csc x}{x^6}$$ Simplify denominator: $$= \frac{-x \csc x \cot x - 3 \csc x}{x^4} = \frac{-\csc x (x \cot x + 3)}{x^4}$$ 3. Differentiate $2x \sin x + \sqrt{x} \cos x$ term-by-term: For $2x \sin x$: product rule $$=2 \sin x + 2x \cos x$$ For $\sqrt{x} \cos x = x^{1/2} \cos x$: product rule $$= \frac{1}{2} x^{-1/2} \cos x + x^{1/2} (-\sin x) = \frac{\cos x}{2 \sqrt{x}} - \sqrt{x} \sin x$$ Total derivative: $$2 \sin x + 2x \cos x + \frac{\cos x}{2 \sqrt{x}} - \sqrt{x} \sin x$$ 4. Differentiate $4 \sec^2 (3x)$ using chain rule: $$=4 \cdot 2 \sec^2 (3x) \tan (3x) \cdot 3 = 24 \sec^2 (3x) \tan (3x)$$ 5. Differentiate $\tan (\sqrt[3]{x}) - \sqrt[3]{\tan x}$: First term: chain rule $$\frac{d}{dx} \tan (x^{1/3}) = \sec^2 (x^{1/3}) \cdot \frac{1}{3} x^{-2/3} = \frac{\sec^2 (x^{1/3})}{3 x^{2/3}}$$ Second term: rewrite as $(\tan x)^{1/3}$, chain rule: $$\frac{d}{dx} (\tan x)^{1/3} = \frac{1}{3} (\tan x)^{-2/3} \cdot \sec^2 x = \frac{\sec^2 x}{3 (\tan x)^{2/3}}$$ Derivative of whole expression: $$\frac{\sec^2 (x^{1/3})}{3 x^{2/3}} - \frac{\sec^2 x}{3 (\tan x)^{2/3}}$$ 6. Differentiate $\cos (x^2) \cos^2 x$ product rule: Let $f=\cos (x^2), g=\cos^2 x$ $$f' = -\sin (x^2) \cdot 2x = -2x \sin (x^2), \quad g' = 2 \cos x (-\sin x) = -2 \cos x \sin x$$ So $$\frac{d}{dx} (f g) = f' g + f g' = -2x \sin (x^2) \cos^2 x - 2 \cos (x^2) \cos x \sin x$$ 7. Differentiate $\sin^3 (\csc 2x)$ chain rule and power rule: Set $h=\sin (\csc 2x)$ then $$\frac{d}{dx} h^3 = 3 h^2 \cdot \frac{d}{dx} h$$ Compute $\frac{d}{dx} h = \cos (\csc 2x) \cdot \frac{d}{dx} (\csc 2x)$ $$= \cos (\csc 2x) \cdot (-\csc 2x \cot 2x) \cdot 2 = -2 \cos (\csc 2x) \csc 2x \cot 2x$$ Final derivative: $$3 \sin^2 (\csc 2x) (-2 \cos (\csc 2x) \csc 2x \cot 2x) = -6 \sin^2 (\csc 2x) \cos (\csc 2x) \csc 2x \cot 2x$$ 8. Differentiate $\cos (\sqrt{\sec (\cot \pi x)})$: chain rule multiple layers Let $$u = \sqrt{\sec (\cot \pi x)} = (\sec (\cot \pi x))^{1/2}$$ $$f= \cos u, f' = -\sin u$$ $$u' = \frac{1}{2} (\sec (\cot \pi x))^{-1/2} \cdot \frac{d}{dx} [\sec (\cot \pi x)]$$ $$= \frac{1}{2 \sqrt{\sec (\cot \pi x)}} \cdot \sec (\cot \pi x) \tan (\cot \pi x) \cdot \frac{d}{dx} (\cot \pi x)$$ $$\frac{d}{dx} (\cot \pi x) = -\csc^2 (\pi x) \cdot \pi = - \pi \csc^2 (\pi x)$$ So $$u' = \frac{1}{2 \sqrt{\sec (\cot \pi x)}} \cdot \sec (\cot \pi x) \tan (\cot \pi x) (- \pi \csc^2 (\pi x))$$ Final derivative: $$-\sin (\sqrt{\sec (\cot \pi x)}) \cdot u' = -\sin (\sqrt{\sec (\cot \pi x)}) \cdot \frac{- \pi \sec (\cot \pi x) \tan (\cot \pi x) \csc^2 (\pi x)}{2 \sqrt{\sec (\cot \pi x)}}$$ Simplify double negative: $$= \frac{\pi \sin (\sqrt{\sec (\cot \pi x)}) \sec (\cot \pi x) \tan (\cot \pi x) \csc^2 (\pi x)}{2 \sqrt{\sec (\cot \pi x)}}$$