Differentiate the following expressions with respect to $x$:

1. Rewrite the radical expression with a fractional exponent to prepare for differentiation.

Let $y = \sqrt[3]{4x - x^2} = (4x - x^2)^{\frac{1}{3}}$

2. Apply the chain rule to differentiate the expression.

$\frac{\text{d}y}{\text{d}x} = \frac{1}{3}(4x - x^2)^{\frac{1}{3}-1} \cdot \frac{\text{d}}{\text{d}x}(4x - x^2)$
$= \frac{1}{3}(4x - x^2)^{-\frac{2}{3}} \cdot (4 - 2x)$

3. Simplify the derivative into a single algebraic fraction.

$\frac{\text{d}y}{\text{d}x} = \frac{4 - 2x}{3(4x - x^2)^{\frac{2}{3}}}$

1. Apply the chain rule to differentiate the numerator.

$\frac{d}{dx}[(3x + 1)^4] = 4(3x + 1)^3 \cdot (3) = 12(3x + 1)^3$

2. Apply the quotient rule to differentiate the given expression.

$\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

$\frac{d}{dx}\left( \frac{(3x+1)^4}{x-3} \right) = \frac{(x - 3)[12(3x + 1)^3] - (3x + 1)^4(1)}{(x - 3)^2}$

3. Factorise the numerator and simplify.

$= \frac{(3x + 1)^3 [12(x - 3) - (3x + 1)]}{(x - 3)^2}$

$= \frac{(3x + 1)^3 (12x - 36 - 3x - 1)}{(x - 3)^2}$

$= \frac{(3x + 1)^3 (9x - 37)}{(x - 3)^2}$

1. Apply the chain rule by differentiating the outer function

$\frac{\text{d}}{\text{d}x} [ \cos(\sin x) ] = -\sin(\sin x) \cdot \frac{\text{d}}{\text{d}x}(\sin x)$

2. Differentiate the inner function to obtain the final expression

$=-\sin(\sin x) \cos x$

1. Apply the standard derivative formula for the inverse sine function along with the chain rule.

$\frac{\text{d}}{\text{d}x} \left[ \sin^{-1}(1 - x) \right] = \frac{1}{\sqrt{1 - (1 - x)^2}} \cdot \frac{\text{d}}{\text{d}x}(1 - x)$
$= \frac{1}{\sqrt{1 - (1 - x)^2}} \cdot (-1)$

2. Expand and simplify the expression under the square root.

$1 - (1 - x)^2 = 1 - (1 - 2x + x^2)$
$= 2x - x^2$

$\therefore \frac{\text{d}}{\text{d}x} \left[ \sin^{-1}(1 - x) \right] = -\frac{1}{\sqrt{2x - x^2}}$

1. Simplify the logarithmic expression using the laws of logarithms.

$\ln \sqrt{\frac{x + 3}{x - 3}} = \ln \left(\frac{x + 3}{x - 3}\right)^{\frac{1}{2}}$
$= \frac{1}{2} \ln \left(\frac{x + 3}{x - 3}\right)$
$= \frac{1}{2} [\ln (x + 3) - \ln (x - 3)]$

2. Differentiate the simplified expression with respect to $x$.

$\frac{\text{d}}{\text{d}x} \left( \frac{1}{2} [\ln (x + 3) - \ln (x - 3)] \right) = \frac{1}{2} \left( \frac{1}{x + 3} - \frac{1}{x - 3} \right)$

3. Combine the fractions over a common denominator and simplify.

$= \frac{1}{2} \left( \frac{(x - 3) - (x + 3)}{(x + 3)(x - 3)} \right)$
$= \frac{1}{2} \left( \frac{-6}{x^2 - 9} \right)$
$= \frac{-3}{x^2 - 9}$

1. Apply the quotient rule to set up the derivative.

$\frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\text{e}^x}{\text{e}^x + \text{e}^{-x}} \right) = \frac{(\text{e}^x + \text{e}^{-x}) \frac{\mathrm{d}}{\mathrm{d}x}(\text{e}^x) - \text{e}^x \frac{\mathrm{d}}{\mathrm{d}x}(\text{e}^x + \text{e}^{-x})}{(\text{e}^x + \text{e}^{-x})^2}$

2. Differentiate the exponential terms and expand the numerator.

$= \frac{(\text{e}^x + \text{e}^{-x})(\text{e}^x) - \text{e}^x(\text{e}^x - \text{e}^{-x})}{(\text{e}^x + \text{e}^{-x})^2}$
$= \frac{\text{e}^{2x} + \text{e}^0 - (\text{e}^{2x} - \text{e}^0)}{(\text{e}^x + \text{e}^{-x})^2}$

3. Simplify the numerator to obtain the final expression.

Since $\text{e}^0 = 1$:
$= \frac{\text{e}^{2x} + 1 - \text{e}^{2x} + 1}{(\text{e}^x + \text{e}^{-x})^2}$
$= \frac{2}{(\text{e}^x + \text{e}^{-x})^2}$