Find the derivatives of each of the following, simplifying your answer.

1. Apply the chain rule to differentiate the function

$\frac{dy}{dx} = 3 \cdot 4\left(1 - \frac{x}{3}\right)^3 \cdot \frac{d}{dx}\left(1 - \frac{x}{3}\right)$

2. Differentiate the inner function and simplify

$\frac{dy}{dx} = 12\left(1 - \frac{x}{3}\right)^3 \cdot \left(-\frac{1}{3}\right)$
$= -4\left(1 - \frac{x}{3}\right)^3$

1. Apply the product rule and chain rule to differentiate the function.

Let $u = 2 - 3x$ and $v = (1 - 4x)^{\frac{1}{2}}$.
$u' = -3$
$v' = \frac{1}{2}(1 - 4x)^{-\frac{1}{2}}(-4) = -2(1 - 4x)^{-\frac{1}{2}}$

$f'(x) = u'v + uv'$
$f'(x) = -3(1 - 4x)^{\frac{1}{2}} + (2 - 3x)\left[-2(1 - 4x)^{-\frac{1}{2}}\right]$

2. Factor out the common term with the negative fractional exponent.

$f'(x) = -3(1 - 4x)^{\frac{1}{2}} - 2(2 - 3x)(1 - 4x)^{-\frac{1}{2}}$
$f'(x) = (1 - 4x)^{-\frac{1}{2}} \left[ -3(1 - 4x) - 2(2 - 3x) \right]$

3. Simplify the expression inside the square brackets to obtain the final answer.

$f'(x) = (1 - 4x)^{-\frac{1}{2}} \left[ -3 + 12x - 4 + 6x \right]$
$f'(x) = (1 - 4x)^{-\frac{1}{2}} (18x - 7)$
$f'(x) = \frac{18x - 7}{\sqrt{1 - 4x}}$

1. Recognise that the derivative of $\frac{dy}{dx}$ is the second derivative $\frac{d^2y}{dx^2}$ and set up the quotient rule

Let $u = 2(3x - 2) = 6x - 4$ and $v = 4 + x$.
Then $\frac{du}{dx} = 6$ and $\frac{dv}{dx} = 1$.
$\frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

2. Apply the quotient rule and simplify the numerator

$\frac{d^2y}{dx^2} = \frac{(4 + x)(6) - (6x - 4)(1)}{(4 + x)^2}$
$= \frac{24 + 6x - 6x + 4}{(4 + x)^2}$
$= \frac{28}{(4 + x)^2}$