1. Differentiate $y = x \sin x$ with respect to $x$. Use the product rule: $\frac{d}{dx}[uv] = u'v + uv'$. Here, $u = x$, $v = \sin x$. So, $u' = 1$, $v' = \cos x$. Therefore, $\frac{dy}{dx} = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x$. 2. Differentiate $y = x^2 e^{2x}$ with respect to $x$. Use the product rule again with $u = x^2$, $v = e^{2x}$. $u' = 2x$, $v' = 2 e^{2x}$ (chain rule). So, $\frac{dy}{dx} = 2x e^{2x} + x^2 \cdot 2 e^{2x} = 2x e^{2x} + 2x^2 e^{2x} = 2x e^{2x}(1 + x)$. 3. Differentiate $y = x^2 \ln x$ with respect to $x$. Again, product rule: $u = x^2$, $v = \ln x$. $u' = 2x$, $v' = \frac{1}{x}$. So, $\frac{dy}{dx} = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x$. 4. Differentiate $y = \sec(ax)$ with respect to $x$. Recall $\frac{d}{dx} \sec u = \sec u \tan u \cdot u'$. Here, $u = ax$, so $u' = a$. Therefore, $\frac{dy}{dx} = \sec(ax) \tan(ax) \cdot a = a \sec(ax) \tan(ax)$. 5. Differentiate $y = \frac{2 \cos 3x}{x^3}$ with respect to $x$. Rewrite as $y = 2 \cos 3x \cdot x^{-3}$. Use product rule: $u = 2 \cos 3x$, $v = x^{-3}$. $u' = 2 \cdot (-\sin 3x) \cdot 3 = -6 \sin 3x$, $v' = -3 x^{-4}$. So, $\frac{dy}{dx} = u'v + uv' = (-6 \sin 3x) x^{-3} + 2 \cos 3x (-3 x^{-4}) = -6 x^{-3} \sin 3x - 6 x^{-4} \cos 3x$. 6. Find the gradient of $y = \frac{2x}{x^2 - 5}$ at $(2, -4)$. Use quotient rule: $y = \frac{f}{g}$, $f=2x$, $g=x^2 - 5$. $f' = 2$, $g' = 2x$. $\frac{dy}{dx} = \frac{f'g - fg'}{g^2} = \frac{2(x^2 - 5) - 2x (2x)}{(x^2 - 5)^2} = \frac{2x^2 - 10 - 4x^2}{(x^2 - 5)^2} = \frac{-2x^2 - 10}{(x^2 - 5)^2}$. Evaluate at $x=2$: Numerator: $-2(2)^2 - 10 = -8 - 10 = -18$. Denominator: $(2^2 - 5)^2 = (4 - 5)^2 = (-1)^2 = 1$. Gradient = $\frac{-18}{1} = -18$. 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. $\frac{dy}{dx} = -6 x^{-3} \sin 3x - 6 x^{-4} \cos 3x$ 6. Gradient at $(2, -4)$ is $-18$.