Last updated on July 19th, 2025
The derivative of ln|x|, which is 1/x, serves as a tool to understand how the logarithmic function changes with respect to variations in x. Derivatives play a crucial role in real-world applications, such as calculating rates of change and optimizing processes. We will now explore the derivative of ln|x| in detail.
The derivative of ln|x| is commonly represented as d/dx (ln|x|) or (ln|x|)', and its value is 1/x. The natural logarithm function ln|x| has a well-defined derivative, indicating it is differentiable within its domain, except at x = 0. The key concepts are mentioned below: Logarithm Function: ln|x| is the natural logarithm of the absolute value of x. Absolute Value: The absolute value function ensures that ln|x| is defined for both positive and negative x. Reciprocal Function: The derivative 1/x represents the reciprocal of x.
The derivative of ln|x| can be denoted as d/dx (ln|x|) or (ln|x|)'. The formula we use to differentiate ln|x| is: d/dx (ln|x|) = 1/x This formula applies to all x ≠ 0.
We can derive the derivative of ln|x| using proofs. To show this, we will use the properties of logarithms and the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using Chain Rule By First Principle The derivative of ln|x| can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of ln|x| using the first principle, consider f(x) = ln|x|. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = ln|x|, we write f(x + h) = ln|x + h|. Substituting these into the equation, f'(x) = limₕ→₀ [ln|x + h| - ln|x|] / h = limₕ→₀ ln(|x + h|/|x|) / h = limₕ→₀ ln(1 + h/x) / h Using the approximation ln(1 + u) ≈ u for small u, f'(x) = limₕ→₀ (h/x) / h = 1/x Thus, f'(x) = 1/x, hence proved. Using Chain Rule To prove the differentiation of ln|x| using the chain rule, Consider the function ln|x| = ln(x) for x > 0 and ln|x| = ln(-x) for x < 0. For x > 0, d/dx (ln(x)) = 1/x. For x < 0, d/dx (ln(-x)) = -1/x x d/dx(-x) = 1/x. Thus, for both positive and negative x, d/dx (ln|x|) = 1/x.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, consider a scenario where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like ln|x|. For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′ (x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues. For the nth derivative of ln|x|, we generally use f⁽ⁿ⁾(x) for the nth derivative of a function f(x), which tells us the change in the rate of change (continuing for higher-order derivatives).
When x is 0, the derivative is undefined because ln|x| is not defined at x = 0. When x is 1, the derivative of ln|1| = 1/1, which is 1.
Students frequently make mistakes when differentiating ln|x|. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of (ln|x|·x³)
Here, we have f(x) = ln|x|·x³. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln|x| and v = x³. Let’s differentiate each term, u′= d/dx (ln|x|) = 1/x v′= d/dx (x³) = 3x² Substituting into the given equation, f'(x) = (1/x)·x³ + ln|x|·3x² Let’s simplify terms to get the final answer, f'(x) = x² + 3x²ln|x| Thus, the derivative of the specified function is x² + 3x²ln|x|.
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.
The height of a water tank is represented by the function y = ln|x|, where y represents the height at a distance x. If x = 2 meters, measure the rate of change of the height.
We have y = ln|x| (height of the tank)...(1) Now, we will differentiate the equation (1). Take the derivative of ln|x|: dy/dx = 1/x Given x = 2, substitute into the derivative: dy/dx = 1/2 Hence, the rate of change of the height at x = 2 is 1/2.
We find the rate of change of the height at x = 2, which indicates the height of the tank changes at half the rate of the horizontal distance.
Derive the second derivative of the function y = ln|x|.
The first step is to find the first derivative, dy/dx = 1/x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx (1/x) = -1/x² Therefore, the second derivative of the function y = ln|x| is -1/x².
We use the step-by-step process, where we start with the first derivative. By differentiating 1/x, we simplify the terms to find the final answer.
Prove: d/dx (ln(x²)) = 2/x.
Let’s start using the chain rule: Consider y = ln(x²) = 2ln(x) To differentiate, we use the chain rule: dy/dx = 2·d/dx (ln(x)) Since the derivative of ln(x) is 1/x, dy/dx = 2·1/x dy/dx = 2/x Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace ln(x) with its derivative. As a final step, we simplify to derive the equation.
Solve: d/dx (ln|x|/x)
To differentiate the function, we use the quotient rule: d/dx (ln|x|/x) = (d/dx (ln|x|)·x - ln|x|·d/dx(x))/x² We substitute d/dx (ln|x|) = 1/x and d/dx (x) = 1 = (1/x·x - ln|x|·1)/x² = (1 - ln|x|)/x² Therefore, d/dx (ln|x|/x) = (1 - ln|x|)/x².
In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result.
Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Logarithm Function: ln|x| is the natural logarithm of the absolute value of x. Absolute Value: The absolute value function ensures that ln|x| is defined for both positive and negative x. Chain Rule: A rule for differentiating compositions of functions, used when differentiating functions like ln|x|. Quotient Rule: A rule for differentiating ratios of functions, applied in finding derivatives of expressions like ln|x|/x.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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