Use logarithm properties to simplify the expression.

$$y = \ln\left(\frac{3x^4}{\sin x}\right) = \ln(3x^4) - \ln(\sin x)$$ $$y = \ln 3 + \ln(x^4) - \ln(\sin x)$$ $$y = \ln 3 + 4\ln x - \ln(\sin x)$$

$$\frac{dy}{dx} = \frac{d}{dx}(\ln 3) + \frac{d}{dx}(4\ln x) - \frac{d}{dx}(\ln(\sin x))$$ $$\frac{dy}{dx} = 0 + 4\left(\frac{1}{x}\right) - \left(\frac{1}{\sin x} \cdot \cos x\right)$$ $$\frac{dy}{dx} = \frac{4}{x} - \frac{\cos x}{\sin x}$$ $$\frac{dy}{dx} = \frac{4}{x} - \cot x$$

The derivative is $\boxed{\frac{4}{x} - \cot x}$.

1. (i) b) A curve is defined parametrically by $x = t^3$, $y = t\left(1 - \frac{1}{4}t^3\right)$. Find the equation of the tangent to the curve at the point where $t = -1$.

$$x_0 = (-1)^3 = -1$$ $$y_0 = (-1)\left(1 - \frac{1}{4}(-1)^3\right) = (-1)\left(1 - \frac{1}{4}(-1)\right) = (-1)\left(1 + \frac{1}{4}\right) = (-1)\left(\frac{5}{4}\right) = -\frac{5}{4}$$

The point is $\left(-1, -\frac{5}{4}\right)$.

$$x = t^3 \implies \frac{dx}{dt} = 3t^2$$ $$y = t - \frac{1}{4}t^4 \implies \frac{dy}{dt} = 1 - \frac{1}{4}(4t^3) = 1 - t^3$$

$$y - \left(-\frac{5}{4}\right) = \frac{2}{3}(x - (-1))$$ $$y + \frac{5}{4} = \frac{2}{3}(x + 1)$$

Multiply by 12:

$$12\left(y + \frac{5}{4}\right) = 12\left(\frac{2}{3}(x + 1)\right)$$ $$12y + 15 = 8(x + 1)$$ $$12y + 15 = 8x + 8$$ $$8x - 12y + 8 - 15 = 0$$ $$8x - 12y - 7 = 0$$

The equation of the tangent is $\boxed{8x - 12y - 7 = 0}$.

1. (ii) a) Resolve $\frac{6}{(1+2x)(3-x)}$ into partial fractions.