Find $f'(x)$ where $f(x) = \ln\bigl(x + \sqrt{1 + x^2}\bigr)$, giving your answer in its simplest form. Find $g'(x)$ where $g(x) = \tan^{-1}\!\left(\dfrac{2 + x}{1 - 2x}\right)$, giving your answer in its simplest form.

1. Differentiate the outer natural logarithm function using the chain rule.

$f'(x) = \frac{1}{x + \sqrt{1 + x^2}} \cdot \frac{d}{dx}\left(x + \sqrt{1 + x^2}\right)$

2. Differentiate the inner expression.

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

3. Substitute the inner derivative back and simplify the full expression.

$\begin{aligned} f'(x) &= \frac{1}{x + \sqrt{1 + x^2}} \left( 1 + \frac{x}{\sqrt{1 + x^2}} \right) \\ &= \frac{1}{x + \sqrt{1 + x^2}} \left( \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}} \right) \\ &= \frac{1}{\sqrt{1 + x^2}} \end{aligned}$

1. Differentiate $g(x)$ using the chain rule and the quotient rule.

$g'(x) = \frac{1}{1 + \left(\frac{2 + x}{1 - 2x}\right)^2} \cdot \frac{(1 - 2x)(1) - (2 + x)(-2)}{(1 - 2x)^2}$

2. Simplify the expression by evaluating the derivative of the quotient.

$g'(x) = \frac{1}{1 + \frac{(2 + x)^2}{(1 - 2x)^2}} \cdot \frac{1 - 2x + 4 + 2x}{(1 - 2x)^2}$
$g'(x) = \frac{1}{\frac{(1 - 2x)^2 + (2 + x)^2}{(1 - 2x)^2}} \cdot \frac{5}{(1 - 2x)^2}$

3. Combine the fractions and expand the denominator to reach the simplest form.

$g'(x) = \frac{(1 - 2x)^2}{(1 - 2x)^2 + (2 + x)^2} \cdot \frac{5}{(1 - 2x)^2}$
$g'(x) = \frac{5}{(1 - 2x)^2 + (2 + x)^2}$
$g'(x) = \frac{5}{1 - 4x + 4x^2 + 4 + 4x + x^2}$
$g'(x) = \frac{5}{5 + 5x^2}$
$g'(x) = \frac{1}{1 + x^2}$