Given the curve equation \(y = x^2 - 2x - 3\), we need to find \(\frac{dy}{dx}\).

Differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = 2x - 2\).

We know \(\frac{dy}{dt} = 4\) and \(\frac{dx}{dt} = 6\).

Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

Substitute the known values:

\(4 = (2x - 2) \cdot 6\).

Solve for \(x\):

\(4 = 12x - 12\)

\(16 = 12x\)

\(x = \frac{16}{12} = \frac{4}{3}\).

(i) The surface area \(S\) of the cuboid is given by:

\(S = 2(x \cdot 2x + 2x \cdot 4x + 4x \cdot x) = 2(2x^2 + 8x^2 + 4x^2) = 28x^2\).

The volume \(V\) of the cuboid is:

\(V = x \cdot 2x \cdot 4x = 8x^3\).

To show \(S = 7V^{\frac{2}{3}}\), substitute \(V = 8x^3\):

\(V^{\frac{2}{3}} = (8x^3)^{\frac{2}{3}} = 4x^2\).

Thus, \(S = 7 \cdot 4x^2 = 28x^2\), which confirms the relation.

(ii) Given \(V = 1000\), find \(x\):

\(8x^3 = 1000 \Rightarrow x^3 = 125 \Rightarrow x = 5\).

\(S = 28x^2 = 28 \cdot 25 = 700\).

\(\frac{dS}{dt} = 2\), \(\frac{dS}{dx} = 56x\), \(\frac{dV}{dx} = 24x^2\).

\(\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}\).

\(\frac{dS}{dt} = \frac{dS}{dx} \cdot \frac{dx}{dt} \Rightarrow 2 = 56x \cdot \frac{dx}{dt} \Rightarrow \frac{dx}{dt} = \frac{1}{140}\).

\(\frac{dV}{dt} = 24x^2 \cdot \frac{1}{140} = \frac{24 \cdot 25}{140} = \frac{30}{7}\).

Thus, the rate of increase of the volume is \(\frac{30}{7}\) or 4.29 cm³ s⁻¹.

Given \(\frac{dy}{dx} = x^3 - \frac{4}{x^2}\), we need to find \(\frac{dy}{dt}\) when \(x = 2\) and \(\frac{dx}{dt} = -0.05\).

First, calculate \(\frac{dy}{dx}\) at \(x = 2\):

\(\frac{dy}{dx} = 2^3 - \frac{4}{2^2} = 8 - 1 = 7\).

Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\).

Substitute the values: \(\frac{dy}{dt} = 7 \times (-0.05) = -0.35\).

Thus, the rate of change of the \(y\)-coordinate is \(-0.35\) units/s.

(i)(a) Differentiate \(y = \frac{1}{2}(4x - 3)^{-1}\) to find \(\frac{dy}{dx} = \left[-\frac{1}{2}(4x - 3)^{-2}\right] \times 4 = -2(4x - 3)^{-2}\).

At \(x = 1\), \(m = -2\).

The gradient of the normal is \(-\frac{1}{m} = \frac{1}{2}\).

The equation of the normal is \(y - \frac{1}{2} = \frac{1}{2}(x - 1)\).

Simplifying gives \(y = \frac{1}{2}x\).

(i)(b) Substitute \(y = \frac{1}{2}x\) into the curve equation: \(\frac{1}{2}(4x - 3)^{-1} = \frac{x}{2}\).

Solving gives \(x = -\frac{1}{4}\).

(ii) Use the chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

At \(x = 1\), \(\frac{dy}{dx} = -2\).

\(\frac{dy}{dt} = (-2) \times (-0.3) = 0.6\).

Given the curve \(y = 2x + \frac{5}{x}\), we need to find \(\frac{dy}{dt}\) when \(x = 1\) and \(\frac{dx}{dt} = 0.02\).

First, differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{d}{dx}(2x + \frac{5}{x}) = 2 - \frac{5}{x^2}\).

Substitute \(x = 1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2 - \frac{5}{1^2} = 2 - 5 = -3\).

Now, use the chain rule to find \(\frac{dy}{dt}\):

\(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

Substitute \(\frac{dy}{dx} = -3\) and \(\frac{dx}{dt} = 0.02\):

\(\frac{dy}{dt} = -3 \times 0.02 = -0.06\).