1. **Problem 1: Differentiate $y = x \sin x$ with respect to $x$.** 2. Use the product rule: If $y = uv$, then $\frac{dy}{dx} = u'v + uv'$. Here, $u = x$, $v = \sin x$. 3. Compute derivatives: $u' = 1$, $v' = \cos x$. 4. Apply product rule: $$\frac{dy}{dx} = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x.$$ 5. **Problem 2: Differentiate $y = x^2 e^{2x}$ with respect to $x$.** 6. Use the product rule again: $u = x^2$, $v = e^{2x}$. 7. Compute derivatives: $u' = 2x$, $v' = 2 e^{2x}$ (chain rule). 8. Apply product rule: $$\frac{dy}{dx} = 2x e^{2x} + x^2 (2 e^{2x}) = 2x e^{2x} + 2x^2 e^{2x} = 2x e^{2x} (1 + x).$$ 9. **Problem 3: Differentiate $y = x^2 \ln x$ with respect to $x$.** 10. Use the product rule: $u = x^2$, $v = \ln x$. 11. Compute derivatives: $u' = 2x$, $v' = \frac{1}{x}$. 12. Apply product rule: $$\frac{dy}{dx} = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x.$$ 13. **Problem 4: Differentiate $y = \sec(ax)$ with respect to $x$.** 14. Recall derivative of $\sec u$ is $\sec u \tan u \cdot u'$. Here, $u = ax$, so $u' = a$. 15. Apply chain rule: $$\frac{dy}{dx} = \sec(ax) \tan(ax) \cdot a = a \sec(ax) \tan(ax).$$ 16. **Problem 5: Differentiate $y = \frac{2 \cos 3x}{x^3}$ with respect to $x$.** 17. Use the quotient rule: If $y = \frac{f}{g}$, then $$\frac{dy}{dx} = \frac{f'g - fg'}{g^2}.$$ Here, $f = 2 \cos 3x$, $g = x^3$. 18. Compute derivatives: $$f' = 2 \cdot (-\sin 3x) \cdot 3 = -6 \sin 3x,$$ $$g' = 3x^2.$$ 19. Apply quotient rule: $$\frac{dy}{dx} = \frac{(-6 \sin 3x) x^3 - (2 \cos 3x)(3x^2)}{x^6} = \frac{-6 x^3 \sin 3x - 6 x^2 \cos 3x}{x^6}.$$ 20. Simplify numerator and denominator: $$= \frac{-6 x^2 (x \sin 3x + \cos 3x)}{x^6} = -6 \frac{x \sin 3x + \cos 3x}{x^4}.$$ 21. **Problem 6: Find the gradient of $y = \frac{2x}{x^2 - 5}$ at point $(2, -4)$.** 22. Use quotient rule with $f = 2x$, $g = x^2 - 5$. 23. Compute derivatives: $$f' = 2,$$ $$g' = 2x.$$ 24. Apply quotient rule: $$\frac{dy}{dx} = \frac{2 (x^2 - 5) - 2x (2x)}{(x^2 - 5)^2} = \frac{2x^2 - 10 - 4x^2}{(x^2 - 5)^2} = \frac{-2x^2 - 10}{(x^2 - 5)^2}.$$ 25. Evaluate at $x=2$: $$\frac{dy}{dx}\bigg|_{x=2} = \frac{-2(2)^2 - 10}{(2^2 - 5)^2} = \frac{-8 - 10}{(4 - 5)^2} = \frac{-18}{(-1)^2} = -18.$$ **Final answers:** 1. $\frac{dy}{dx} = \sin x + x \cos x$ 2. $\frac{dy}{dx} = 2x e^{2x} (1 + x)$ 3. $\frac{dy}{dx} = 2x \ln x + x$ 4. $\frac{dy}{dx} = a \sec(ax) \tan(ax)$ 5. $\frac{dy}{dx} = -6 \frac{x \sin 3x + \cos 3x}{x^4}$ 6. Gradient at $(2, -4)$ is $-18$.