Problem 70 - Calculus Early Transcendentals

Question

Differentiate \(\ln x\) for \(x>0\) and differentiate \(\ln (-x)\) for \(x<0\) to conclude that \(\frac{d}{d x}(\ln |x|)=\frac{1}{x}\).

Step-by-Step Solution

Answer

The derivative is \(\frac{1}{x}\).

Step 1: Differentiate \(\ln x\) for \(x > 0\)

To differentiate \(\ln x\) for \(x > 0\), we will use the basic definition of the natural logarithm. This definition is as follows: $$ \frac{d}{dx}(\ln x) = \frac{1}{x} $$ So in this case, the derivative of \(\ln x\) with respect to x is \(\frac{1}{x}\).

Step 2: Differentiate \(\ln(-x)\) for \(x < 0\)

For differentiating \(\ln(-x)\) with respect to x for \(x < 0\), we will use the chain rule. This is because the natural logarithm function is again taking another function, \(-x\), as its input. The chain rule states: $$ \frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) $$ Using this rule, we have the following \(f(g(x))\) definition: $$ f(g(x)) = \ln(-x) \\ f'(g(x)) = \frac{1}{-x} \\ g(x) = -x \\ g'(x) = -1 $$ Applying the chain rule: $$ \frac{d}{dx}(\ln(-x)) = \frac{1}{-x} \cdot (-1) $$ So the derivative of \(\ln(-x)\) with respect to x is \(\frac{1}{x}\), as the negative signs cancel out.

Step 3: Deduce the derivative of \(\ln|x|\)

Now that we have both derivatives, we can find the derivative of \(\ln|x|\) as follows: For \(x > 0\), \(\ln|x| = \ln x\), and the derivative of \(\ln x\) is \(\frac{1}{x}\). For \(x < 0\), \(\ln|x| = \ln(-x)\), and the derivative of \(\ln(-x)\) is also \(\frac{1}{x}\). Together, we have the following result: $$ \frac{d}{dx}(\ln|x|) = \frac{1}{x} $$ This shows that the derivative of \(\ln|x|\) with respect to x is \(\frac{1}{x}\) for all \(x \neq 0\).

Key Concepts

Natural Logarithm

The natural logarithm, represented by \(\ln x\), is one of the most important functions in calculus. It appears frequently in mathematical equations and real-world applications.

• The natural logarithm is \(\ln x\), which is the logarithm to the base \(e\). Here, \("e"\) is a mathematical constant approximately equal to 2.71828.
• It is important because it simplifies many calculations, especially when dealing with growth and decay problems.
• In calculus, the derivative of the natural logarithm \(\ln x\) is crucial because it leads to further understanding of growth rates and changes over time.

For positive values of \(x\), the natural logarithm is defined, and its derivative is \(\frac{1}{x}\). This property is key in connecting differential calculus with real-world scenarios.
When considering negative inputs \(x < 0\), we do not use \(\ln x\) directly, but use a related expression, like \(\ln(-x)\) or \(\ln|x|\). This handling is important as it allows calculation of derivatives even for negative inputs by respecting the nature of logarithms.

Derivative

Derivatives are fundamental to understanding rates of change. In calculus, finding the derivative of a function involves determining how the function value changes as its input changes slightly.

• The derivative of a function at a given point informs us of the behavior of the function around that point.
• For \(\ln x\), the derivative is straightforward: \(\frac{d}{dx}(\ln x) = \frac{1}{x}\) for \(x > 0\). This derivative shows how \(\ln x\) changes with respect to changes in \(x\).
• The derivative \(\frac{1}{x}\) becomes critical when dealing with problems that involve natural logarithms.

Differentiating \(\ln (-x)\) involves using a derivative rule known as the chain rule, which unveils another layer of understanding as we seek the derivative for negative inputs. Recognizing that \(\ln|x|\) has the same derivative as \(\ln x\) (\frac{1}{x}), across all \(x eq 0\), is insightful because it shows the symmetry and continuity of the logarithmic function in calculus.

Chain Rule

The chain rule is an essential tool in differential calculus, especially when dealing with functions composed of other functions. It simplifies the process of finding derivatives of complex expressions.

• The chain rule states that the derivative of a composition of functions \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\).
• This is particularly useful when differentiating functions like \(\ln(-x)\).

In this scenario, when differentiating \(\ln(-x)\):

• Consider \(g(x) = -x\), which indicates an input transformation.
• The derivative \(-x\) is \(-1\), which is part of the multiplication process in the chain rule.
• Inside the natural log function, \(f(x) = \ln x\), the derivative is \(\frac{1}{x}\).

Applying the chain rule: \(\frac{d}{dx}(\ln(-x)) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}\). The negative signs cancel, revealing that the derivative of \(\ln(-x)\) results in the same \(\frac{1}{x}\), just as with \(\ln x\), aligning with how derivatives should treat both positive and negative input transformations similarly.