1. **Problem (a): Find the area enclosed by the curve $y = 4\cos 3x$, the x-axis, and the lines $x=0$ and $x=\frac{\pi}{6}$.** The area under a curve from $x=a$ to $x=b$ is given by the definite integral: $$\text{Area} = \int_a^b y \, dx$$ Here, $y = 4\cos 3x$, $a=0$, and $b=\frac{\pi}{6}$. Calculate the integral: $$\int_0^{\frac{\pi}{6}} 4\cos 3x \, dx = 4 \int_0^{\frac{\pi}{6}} \cos 3x \, dx$$ Use substitution or recall that: $$\int \cos kx \, dx = \frac{1}{k} \sin kx + C$$ So, $$4 \int_0^{\frac{\pi}{6}} \cos 3x \, dx = 4 \left[ \frac{1}{3} \sin 3x \right]_0^{\frac{\pi}{6}} = \frac{4}{3} \left( \sin \frac{3\pi}{6} - \sin 0 \right)$$ Evaluate the sine values: $$\sin \frac{3\pi}{6} = \sin \frac{\pi}{2} = 1, \quad \sin 0 = 0$$ Therefore, $$\text{Area} = \frac{4}{3} (1 - 0) = \frac{4}{3}$$ --- 2. **Problem (b): Differentiate $y = e^{3t} \sin 4t$ with respect to $t$.** Use the product rule: $$\frac{dy}{dt} = \frac{d}{dt} (e^{3t}) \cdot \sin 4t + e^{3t} \cdot \frac{d}{dt} (\sin 4t)$$ Calculate each derivative: $$\frac{d}{dt} (e^{3t}) = 3 e^{3t}$$ $$\frac{d}{dt} (\sin 4t) = 4 \cos 4t$$ Substitute back: $$\frac{dy}{dt} = 3 e^{3t} \sin 4t + 4 e^{3t} \cos 4t = e^{3t} (3 \sin 4t + 4 \cos 4t)$$ --- 3. **Problem (c): Determine the differential coefficient of $y = \tan ax$.** Recall the derivative of $\tan u$ with respect to $x$ is: $$\frac{d}{dx} (\tan u) = \sec^2 u \cdot \frac{du}{dx}$$ Here, $u = ax$, so: $$\frac{du}{dx} = a$$ Therefore, $$\frac{dy}{dx} = a \sec^2 (ax)$$ --- **Final answers:** (a) Area = $\frac{4}{3}$ (b) $\frac{dy}{dt} = e^{3t} (3 \sin 4t + 4 \cos 4t)$ (c) $\frac{dy}{dx} = a \sec^2 (ax)$