# What is the derivative of 1/ln x?

#### Understand the Problem

The question is asking for the derivative of the function 1/ln(x). To solve this, we will apply the quotient rule of differentiation, since it involves a division of two functions.

$-\frac{1}{x [\ln(x)]^2}$

The final answer is $-\frac{1}{x [\ln(x)]^2}$

#### Steps to Solve

1. Rewrite the function for easier differentiation

Rewrite the function using a negative exponent:

$$f(x) = \frac{1}{\ln(x)} = [\ln(x)]^{-1}$$

This makes it easier to apply the chain rule.

1. Apply the chain rule

The chain rule states that the derivative of $[u(x)]^n$ is $n[u(x)]^{n-1} , u'(x)$. Here, $u(x) = \ln(x)$ and $n = -1$:

$$\frac{d}{dx}[\ln(x)]^{-1} = -1 [\ln(x)]^{-2} \frac{d}{dx}[\ln(x)]$$

1. Find the derivative of $\ln(x)$

The derivative of $\ln(x)$ with respect to $x$ is $\frac{1}{x}$:

$$\frac{d}{dx}[\ln(x)] = \frac{1}{x}$$

1. Combine the results

Substitute the derivative of $\ln(x)$ back into the previous result:

$$\frac{d}{dx}[\ln(x)]^{-1} = -1 [\ln(x)]^{-2} \cdot \frac{1}{x}$$

Simplify the expression:

$$\frac{d}{dx}\left( \frac{1}{\ln(x)} \right) = -\frac{1}{x[\ln(x)]^2}$$

The final answer is $-\frac{1}{x [\ln(x)]^2}$

This form allows us to differentiate a composition of functions more easily. The result shows that the rate of change of $\frac{1}{\ln(x)}$ decreases as x increases.
A common mistake is to forget to apply the chain rule properly or to incorrectly differentiate $\ln(x)$. Always remember that the derivative of $\ln(x)$ is $\frac{1}{x}$.