The question asks for the formula for the derivative of the division of two functions, $f(x)$ and $g(x)$, where:

$f(x) = x + 1 \quad \text{and} \quad g(x) = x^2 - 1.$

To find $\frac{f}{g}'(x)$, we use the quotient rule for differentiation:

$\left( \frac{f}{g} \right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}.$

Steps:

1. Find $f'(x)$: $f'(x) = \frac{d}{dx}(x + 1) = 1.$

2. Find $g'(x)$: $g'(x) = \frac{d}{dx}(x^2 - 1) = 2x.$

3. Substitute $f(x), f'(x), g(x), g'(x)$ into the quotient rule: $\left( \frac{f}{g} \right)'(x) = \frac{(1)(x^2 - 1) - (x + 1)(2x)}{(x^2 - 1)^2}.$

4. Simplify the numerator: $(1)(x^2 - 1) - (x + 1)(2x) = x^2 - 1 - 2x^2 - 2x = -x^2 - 2x - 1.$

5. Final formula: $\left( \frac{f}{g} \right)'(x) = \frac{-x^2 - 2x - 1}{(x^2 - 1)^2}.$

The derivative of $\frac{f(x)}{g(x)}$ is:

$\boxed{\frac{-x^2 - 2x - 1}{(x^2 - 1)^2}}$

Would you like a more detailed explanation of any step? Here are five related questions to consider:

1. How is the quotient rule applied to general functions $f(x)$ and $g(x)$?
2. Why is it necessary to calculate both $f'(x)$ and $g'(x)$?
3. How do we simplify the numerator in the quotient rule?
4. Can this formula be further factorized or simplified?
5. What happens if $g(x) = 0$ at certain points?

Tip: Always double-check your simplifications, especially in the numerator, to ensure the final formula is correct!