Differentiate the following expressions with respect to x:

1. Express the radical term with a fractional index to prepare for differentiation.

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

2. Apply the general power rule (chain rule) to differentiate the expression.

$\dfrac{dy}{dx} = \dfrac{1}{3}(4x - x^{2})^{\frac{1}{3}-1} \cdot \dfrac{d}{dx}(4x - x^{2})$

3. Evaluate the derivative of the inner function and state the simplified final answer.

$\dfrac{dy}{dx} = \dfrac{1}{3}(4x - x^{2})^{-\frac{2}{3}} (4 - 2x)$
$= \dfrac{4 - 2x}{3(4x - x^2)^{\frac{2}{3}}}$

1. Apply the quotient rule and chain rule to differentiate the expression.

Let $y = \dfrac{(3x+1)^4}{x-3}$.

Using the quotient rule $\dfrac{d}{dx}\left[\dfrac{u}{v}\right] = \dfrac{v u' - u v'}{v^2}$:
$\frac{dy}{dx} = \frac{(x-3) \frac{d}{dx}[(3x+1)^4] - (3x+1)^4 \frac{d}{dx}[x-3]}{(x-3)^2}$

Applying the chain rule for the term $(3x+1)^4$:
$\frac{dy}{dx} = \frac{(x-3)[4(3x+1)^3(3)] - (3x+1)^4(1)}{(x-3)^2}$
$= \frac{12(x-3)(3x+1)^3 - (3x+1)^4}{(x-3)^2}$

2. Factorise and simplify the numerator to obtain the final expression.

Factor out the common term $(3x+1)^3$ from the numerator:
$\frac{dy}{dx} = \frac{(3x+1)^3 [12(x-3) - (3x+1)]}{(x-3)^2}$
$= \frac{(3x+1)^3 (12x - 36 - 3x - 1)}{(x-3)^2}$
$= \frac{(3x+1)^3 (9x - 37)}{(x-3)^2}$

1. Apply the chain rule by differentiating the outer cosine function

Let $y = \cos(\sin x)$. Using the chain rule, we differentiate the outer $\cos$ function first, leaving the inner argument unchanged:
$\frac{dy}{dx} = -\sin(\sin x) \cdot \frac{d}{dx}(\sin x)$

2. Differentiate the inner sine function and simplify

Differentiate the inner function $\sin x$ to get $\cos x$, then multiply:
$\frac{dy}{dx} = -\sin(\sin x) \cdot \cos x$
$\frac{dy}{dx} = -\cos x \sin(\sin x)$

1. Apply the chain rule using the standard derivative formula for inverse sine.

Let $y = \sin^{-1}(1-x)$.
Using the chain rule $\frac{d}{dx}[\sin^{-1}(u)] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$,
$\frac{dy}{dx} = \frac{1}{\sqrt{1 - (1-x)^2}} \cdot \frac{d}{dx}(1-x)$
$= \frac{-1}{\sqrt{1 - (1-x)^2}}$

2. Simplify the algebraic expression inside the square root.

$\frac{dy}{dx} = \frac{-1}{\sqrt{1 - (1 - 2x + x^2)}}$
$= \frac{-1}{\sqrt{2x - x^2}}$

1. Simplify the logarithmic expression using the laws of logarithms to make differentiation easier.

Let $y = \ln\sqrt{\dfrac{x+3}{x-3}}$.

$\begin{aligned} y &= \frac{1}{2} \ln\left(\dfrac{x+3}{x-3}\right) \\ &= \frac{1}{2} [\ln(x+3) - \ln(x-3)] \end{aligned}$

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

$\begin{aligned} \frac{dy}{dx} &= \frac{1}{2} \left[ \frac{d}{dx}\ln(x+3) - \frac{d}{dx}\ln(x-3) \right] \\ &= \frac{1}{2} \left( \dfrac{1}{x+3} - \dfrac{1}{x-3} \right) \end{aligned}$

3. Combine the fractions into a single algebraic term.

$\begin{aligned} \frac{dy}{dx} &= \frac{1}{2} \left( \dfrac{(x-3) - (x+3)}{(x+3)(x-3)} \right) \\ &= \frac{1}{2} \left( \dfrac{-6}{x^2-9} \right) \\ &= -\dfrac{3}{x^2-9} \end{aligned}$

1. Apply the quotient rule to the expression

$\frac{d}{dx} \left[ \frac{e^x}{e^x + e^{-x}} \right] = \frac{(e^x + e^{-x}) \frac{d}{dx}(e^x) - e^x \frac{d}{dx}(e^x + e^{-x})}{(e^x + e^{-x})^2}$

2. Evaluate the derivatives of the exponential terms in the numerator

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

3. Expand the terms in the numerator using laws of indices

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

4. Simplify the numerator to obtain the final expression