The problem has three parts: (i) Differentiate $y = 2x^3 \sin 2x$ with respect to $x$. (ii) Simplify $(\frac{8}{27})^{-2/3}$. (iii) Express $\frac{2+i}{-1+3i}$ in the standard form $a+bi$.

1. Problem Description

The problem has three parts:

(i) Differentiate $y = 2x^3 \sin 2x$ with respect to $x$.

(ii) Simplify $(\frac{8}{27})^{-2/3}$.

(iii) Express $\frac{2+i}{-1+3i}$ in the standard form $a+bi$.

2. Solution Steps

(i) To differentiate $y = 2x^3 \sin 2x$ with respect to $x$, we use the product rule:

If $y = uv$, then $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$.

Here, $u = 2x^3$ and $v = \sin 2x$.

$\frac{du}{dx} = 6x^2$.

$\frac{dv}{dx} = 2 \cos 2x$.

Therefore,

$\frac{dy}{dx} = (2x^3)(2\cos 2x) + (\sin 2x)(6x^2)$

$\frac{dy}{dx} = 4x^3 \cos 2x + 6x^2 \sin 2x$

(ii) To simplify $(\frac{8}{27})^{-2/3}$, we have

$(\frac{8}{27})^{-2/3} = (\frac{27}{8})^{2/3}$

$= (\sqrt[3]{\frac{27}{8}})^2$

$= (\frac{\sqrt[3]{27}}{\sqrt[3]{8}})^2$

$= (\frac{3}{2})^2$

$= \frac{9}{4}$

(iii) To express $\frac{2+i}{-1+3i}$ in the standard form $a+bi$, we multiply the numerator and denominator by the conjugate of the denominator, which is $-1-3i$.

$\frac{2+i}{-1+3i} = \frac{(2+i)(-1-3i)}{(-1+3i)(-1-3i)}$

$= \frac{-2 - 6i - i - 3i^2}{1 + 3i - 3i - 9i^2}$

$= \frac{-2 - 7i - 3(-1)}{1 - 9(-1)}$

$= \frac{-2 - 7i + 3}{1 + 9}$

$= \frac{1 - 7i}{10}$

$= \frac{1}{10} - \frac{7}{10}i$

3. Final Answer

(i) $\frac{dy}{dx} = 4x^3 \cos 2x + 6x^2 \sin 2x$

(ii) $\frac{9}{4}$

(iii) $\frac{1}{10} - \frac{7}{10}i$