Find the derivative of f(x) = -3x2/x3 using first principles, and also use differentiation rules for y = 4e2x + 1/2 ln x - 3/√x + sin 30°.

The derivative of $f(x)$ by first principles is $\boxed{\frac{3}{x^2}}$.

3.2
Use differentiation rules to determine the derivatives of each of the following:

3.2.1
Given $y = 4e^{2x} + \frac{1}{2}\ln x - \frac{3}{\sqrt{x}} + \sin 30^\circ$.

Remember $\sin 30^\circ$ is a constant.

The derivative is $\boxed{8e^{2x} + \frac{1}{2x} + \frac{3}{2\sqrt{x^3}}}$.

3.2.2
Given $y = \frac{6x^4 - 3x^2}{x^2}$.

The derivative is $\boxed{12x}$.

3.2.3
Given $y = 3x^3 \sin(2x)$.

$\frac{d}{dx}(uv) = u'v + uv'$.
Let $u = 3x^3$ and $v = \sin(2x)$.
Then $u' = 9x^2$ and $v' = 2\cos(2x)$.

The derivative is $\boxed{9x^2 \sin(2x) + 6x^3 \cos(2x)}$.

3.2.4
Given $y = \frac{3}{(2-x)^4}$.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
Let $u = 2-x$, so $\frac{du}{dx} = -1$.
Then $y = 3u^{-4}$, so $\frac{dy}{du} = 3(-4)u^{-5} = -12u^{-5}$.

The derivative is $\boxed{\frac{12}{(2-x)^5}}$.

3.2.5
Given $y = \ln[\cos(3x)]$.

Let $u = \cos(3x)$, so $\frac{du}{dx} = -3\sin(3x)$.
Then $y = \ln u$, so $\frac{dy}{du} = \frac{1}{u}$.

The derivative is $\boxed{-3\tan(3x)}$.

3.2.6
Given $y = -3e^{5x} + \frac{2}{5x^{-2}} - \sqrt[3]{x^5} + \frac{\pi}{2}$.

Remember $\frac{\pi}{2}$ is a constant.