To find the derivative of the function $f(x) = \frac{1}{\ln(x)}$, we can use the chain rule and quotient rule.

Steps:

1. Write the function in a more convenient form:
   $f(x) = (\ln(x))^{-1}$

2. Apply the chain rule:
   The derivative of $(\ln(x))^{-1}$ is: $f'(x) = -(\ln(x))^{-2} \cdot \frac{d}{dx}[\ln(x)]$

3. The derivative of $\ln(x)$ is $\frac{1}{x}$, so: $f'(x) = -\frac{1}{x} \cdot (\ln(x))^{-2}$

4. Simplify the expression: $f'(x) = -\frac{1}{x (\ln(x))^2}$

Final Answer:

$f'(x) = -\frac{1}{x (\ln(x))^2}$

Would you like more details on this derivation, or do you have any questions?

Related Questions:

1. What is the derivative of $\ln(x^2 + 1)$?
2. How do you differentiate $x \ln(x)$?
3. What is the second derivative of $\frac{1}{\ln(x)}$?
4. Can you integrate $\frac{1}{\ln(x)}$?
5. How would you differentiate $\ln(\sin(x))$?

Tip:

When dealing with derivatives of logarithmic functions, always check if the chain rule applies for nested functions.