Question

For the following exercises, find the derivative \(\frac{d y}{d x}\). $$ y=\frac{1}{\ln x} $$

Short answer

The derivative is \( \frac{dy}{dx} = -\frac{1}{x(\ln x)^2} \).

Step-by-step solution

Identify the Function Type

The given function is a reciprocal of a logarithmic function: \( y = \frac{1}{\ln x} \). This can be rewritten as \( y = (\ln x)^{-1} \).

Apply the Chain Rule

To find the derivative, use the chain rule. If \( u = \ln x \), then \( y = u^{-1} \) where \( y = (\ln x)^{-1} \). The derivative \( \frac{dy}{dx} \) is computed as the derivative of \( u^{-1} \) with respect to \( u \), multiplied by the derivative of \( u \) with respect to \( x \).

Derive the Outer Function

The derivative of \( u^{-1} \) with respect to \( u \) is \( -u^{-2} \). So, \( \frac{d}{du}[u^{-1}] = -u^{-2} \).

Derive the Inner Function

The derivative of \( u = \ln x \) with respect to \( x \) is \( \frac{1}{x} \). So, \( \frac{du}{dx} = \frac{1}{x} \).

Apply the Chain Rule to Complete the Derivative

Combine the derivatives from the outer and inner functions using the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -u^{-2} \cdot \frac{1}{x} = -\frac{1}{(\ln x)^2} \cdot \frac{1}{x} \).

Simplify the Expression

Simplify the expression to find the derivative: \( \frac{dy}{dx} = -\frac{1}{x(\ln x)^2} \).

Key concepts

Derivative of Logarithmic Functions

Logarithmic functions, such as \(\ln x\), are a common part of calculus. They are the inverse of exponential functions. To find their derivatives, it's helpful to know their nature. The derivative of \(\ln x\) itself is relatively simple: \(\frac{1}{x}\).

When combined with other rules of differentiation, like the chain rule or product rule, derivatives for more complex expressions involving logarithms can be found. In the given exercise, you are dealing with the reciprocal of a logarithm, \(y = \frac{1}{\ln x}.\)

By recognizing the logarithmic part in this expression, you can efficiently apply different differentiation rules. This particular function is a good example of how important understanding the derivative of \(\ln x\) is to broader applications.

Reciprocal Functions

Reciprocal functions take the form \(\begin{align*}\frac{1}{f(x)}\end{align*}\) and are quite common in calculus problems. These forms require careful consideration when applying calculus rules. The reciprocal of a logarithmic function, \(y = \frac{1}{\ln x},\) means you inherently have a function raised to the power of -1.

To differentiate such functions, you can rewrite them using the negative exponent meaning: \(y = (\ln x)^{-1}.\) Recognizing this form allows the derivative to be computed by applying the chain rule. The chain rule simplifies differentiating functions that are compositions of other functions.

For \(y = (\ln x)^{-1},\) you first derive the outer function based on the power rule, then multiply by the derivative of the inner function, \(\ln x.\) This method highlights the efficiency and importance of recognizing and rewritten forms in calculus.

Simplifying Derivatives

Once you apply differentiation rules, it's essential to simplify the resulting expression. Simplification is the process of reducing a derivative to its most straightforward form. In the given exercise, after using the chain rule, the expression for the derivative becomes:

• -\(\frac{1}{(\ln x)^2}\) from the outer derivative
• \(\frac{1}{x}\) from the inner derivative

Combining these results using multiplication yields: \(\frac{dy}{dx} = -\frac{1}{x(\ln x)^2}.\)

This final simplification is essential. It makes evaluating the derivative much faster and more manageable, particularly for further calculations or analysis. Simplifying derivatives not only enhances the computational ease but also provides better insight into the behavior of the function over its domain.