Problem 297 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Problem Type

The problem involves finding the derivative of the function \( y = \frac{1}{\ln x} \). This is a quotient, and we will apply the derivative rules accordingly.

Step 2: Express in Terms of Powers

Rewrite \( y = \frac{1}{\ln x} \) as \( y = (\ln x)^{-1} \) to simplify the differentiation process by avoiding the quotient rule.

Step 3: Apply the Chain Rule

Use the chain rule to differentiate \( y = (\ln x)^{-1} \). The chain rule formula is \( \frac{d}{dx}[u^n] = n \cdot u^{n-1} \cdot \frac{du}{dx} \), where \( u = \ln x \) and \( n = -1 \).

Step 4: Differentiate the Outer Function

Differentiate the outer function \( u^n = (\ln x)^{-1} \) using \( n \cdot u^{n-1} = -1 \cdot (\ln x)^{-2} = - (\ln x)^{-2} \).

Step 5: Differentiate the Inner Function

Differentiate the inner function \( u = \ln x \). The derivative of \( \ln x \) is \( \frac{d}{d x}(\ln x) = \frac{1}{x} \).

Step 6: Combine the Derivatives

Multiply the results from the outer and inner derivatives. Thus, \( \frac{d y}{d x} = - (\ln x)^{-2} \cdot \frac{1}{x} = -\frac{1}{x (\ln x)^2} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of composite functions. If you have a function that is made by combining two or more functions, the chain rule helps you differentiate it.
For example, consider a function in the form of \( y = f(g(x)) \). To find its derivative with respect to \( x\), you use the rule: \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \).
This means you differentiate the outer function and multiply it by the derivative of the inner function.

• The outer derivative focuses on the overall exponent or highest structure of the function.
• The inner derivative focuses on the transformation within the function, like a logarithm or another layer of math.

In our example, for \( y = (\ln x)^{-1} \), the chain rule breaks it into two parts: differentiating the power function and the natural logarithm function. Doing this ensures we capture every aspect of the composite structure.

Quotient Rule

The quotient rule is used for differentiating functions that are expressed as a division of two other functions, basically a "top over bottom" setup. However, you might notice we didn't use it here directly.
In calculus, the quotient rule is formally given by: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \]

• \( u \) and \( v \) are functions of \( x \).
• \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) are the derivatives of \( u \) and \( v \) respectively.

Using the formula can sometimes be complex, especially when the functions have power terms or logs themselves. A neat trick is to rewrite our original function using negative exponents (which we did here), thereby simplifying the process and avoiding potentially lengthy calculations while still respecting the integrity of the original functions.

Logarithmic Differentiation

Logarithmic differentiation is an awesome trick for differentiating products, quotients, or powers of functions, especially when logs are involved. It's particularly helpful when directly differentiating might get a bit hairy.
The idea is to take the natural log on both sides of the function, which transforms multiplication into addition and powers into coefficients, greatly simplifying differentiation.
For instance, with \( y = \frac{1}{\ln x} \), taking the log simplifies our understanding and lets calculus rules like the power rule do some heavy lifting.

• First, we reframe the function to a power form, \( y = (\ln x)^{-1} \).
• Then, differentiate as if dealing with a simpler form, and finally apply the properties of logs.

This method was not strictly needed in our example, but knowing it broadens our toolbox and prepares us for more complex derivatives.