The given function is:

$f(t) = \frac{9t}{t^3 - t - 1}$

We need to differentiate this function $f(t)$ with respect to $t$. This is a quotient of two functions, so we will apply the quotient rule for differentiation.

The quotient rule is given by:

$\frac{d}{dt} \left( \frac{u(t)}{v(t)} \right) = \frac{v(t) u'(t) - u(t) v'(t)}{(v(t))^2}$

where $u(t) = 9t$ and $v(t) = t^3 - t - 1$.

Step 1: Differentiate $u(t) = 9t$

$u'(t) = 9$

Step 2: Differentiate $v(t) = t^3 - t - 1$

$v'(t) = 3t^2 - 1$

Step 3: Apply the quotient rule

Now applying the quotient rule:

$f'(t) = \frac{(t^3 - t - 1)(9) - (9t)(3t^2 - 1)}{(t^3 - t - 1)^2}$

Simplify the numerator:

$f'(t) = \frac{9(t^3 - t - 1) - 9t(3t^2 - 1)}{(t^3 - t - 1)^2}$

Expanding both terms:

$f'(t) = \frac{9t^3 - 9t - 9 - 9t(3t^2 - 1)}{(t^3 - t - 1)^2}$ $= \frac{9t^3 - 9t - 9 - (27t^3 - 9t)}{(t^3 - t - 1)^2}$ $= \frac{9t^3 - 9t - 9 - 27t^3 + 9t}{(t^3 - t - 1)^2}$ $= \frac{-18t^3 - 9}{(t^3 - t - 1)^2}$

Thus, the derivative is:

$f'(t) = \frac{-18t^3 - 9}{(t^3 - t - 1)^2}$

Would you like further details or have any questions?

Here are 5 related questions you can explore:

1. How does the quotient rule differ from the product rule?
2. Can you differentiate a similar function, such as $f(x) = \frac{5x}{x^2 + 3x - 4}$?
3. What is the importance of simplifying terms after using the quotient rule?
4. How would this differentiation process change if the function had more complex terms in the denominator?
5. What happens to the function if the denominator becomes zero?

Tip: Always simplify your expression as much as possible after applying the quotient or product rule to avoid errors.