Derivative of ln|x| Differentiate ln x, for x \u003e 0, and differentiate ln(−x), for x \u003c 0, to conclude that d/dx (ln|x|) = 1/x

Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

문제 7.1.72

7장, 문제 7.1.72

Derivative of ln|x| Differentiate ln x, for x > 0, and differentiate ln(−x), for x < 0, to conclude that d/dx (ln|x|) = 1/x

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Recall that the function \( \ln|x| \) can be expressed piecewise as \( \ln x \) when \( x > 0 \) and \( \ln(-x) \) when \( x < 0 \).

For \( x > 0 \), differentiate \( \ln x \) using the derivative rule for natural logarithm: \( \frac{d}{dx} \ln x = \frac{1}{x} \).

For \( x < 0 \), rewrite \( \ln|x| = \ln(-x) \). Use the chain rule to differentiate: \( \frac{d}{dx} \ln(-x) = \frac{1}{-x} \cdot \frac{d}{dx}(-x) \).

Calculate the derivative inside the chain rule: \( \frac{d}{dx}(-x) = -1 \), so the derivative becomes \( \frac{1}{-x} \times (-1) = \frac{1}{x} \).

Combine both cases to conclude that for all \( x \neq 0 \), \( \frac{d}{dx} \ln|x| = \frac{1}{x} \).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Derivative of the Natural Logarithm Function

The derivative of ln(x) for x > 0 is 1/x. This follows from the definition of the natural logarithm as the inverse of the exponential function and the chain rule, reflecting how the rate of change of ln(x) depends inversely on x.

Chain Rule for Differentiation

The chain rule allows differentiation of composite functions. For ln(−x) when x < 0, we treat −x as an inner function, differentiating ln(u) with respect to u and then u = −x with respect to x, resulting in the derivative −1/x.

Absolute Value Function and Piecewise Differentiation

The function ln|x| is defined piecewise as ln(x) for x > 0 and ln(−x) for x < 0. Differentiating each piece separately and combining results shows that d/dx (ln|x|) = 1/x for all x ≠ 0, capturing the behavior on both sides of zero.