A) Differentiate ((3x3)/( x)) with respect to x. b) Find the equation of the tangent to the curve defined parametrically by x = t3, y = t(1 − (1)/(4)t3) at the point where t = -1.

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

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

The derivative is $\boxed{\frac{3}{x} - \cot x}$.

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

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

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

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