Last updated on July 19th, 2025
We use the derivative of ln(x), which is 1/x, as a measuring tool for how the natural logarithm function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of ln(x) in detail.
We now understand the derivative of ln x. It is commonly represented as d/dx (ln x) or (ln x)', and its value is 1/x. The function ln x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Natural Logarithm Function: ln(x) is the natural logarithm of x. Quotient Rule: Not needed for ln(x), but useful for functions involving divisions. Reciprocal Function: The derivative results in a reciprocal function, 1/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 The formula applies to all x > 0, as ln(x) is defined only for positive x.
We can derive the derivative of ln x using proofs. To show this, we will use the definition of the derivative along with the rules of differentiation. Here are the methods we use to prove this: 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 … (1) Given that f(x) = ln x, we write f(x + h) = ln(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [ln(x + h) - ln x] / h Using the property of logarithms, ln(a) - ln(b) = ln(a/b), f'(x) = 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 Hence, proved. Using Chain Rule To prove the differentiation of ln x using the chain rule, Consider ln x as u, where x = e^u. Then, d/dx (ln x) = d/du (e^u) · du/dx Since d/du (e^u) = e^u, and du/dx = 1/x, d/dx (ln x) = e^u / x = 1/x Using Logarithmic Differentiation Let y = ln x. Taking the exponential of both sides, e^y = x. Differentiating both sides with respect to x, d/dx (e^y) = d/dx (x) e^y · dy/dx = 1 Substituting back y = ln x, we get, x · dy/dx = 1 dy/dx = 1/x Therefore, the derivative of ln x is 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, think of a car 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 fⁿ(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 for x ≤ 0. When x is 1, the derivative of ln x = 1/x, 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 · x²)
Here, we have f(x) = ln x · x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln x and v = x². Let’s differentiate each term, u′ = d/dx (ln x) = 1/x v′ = d/dx (x²) = 2x Substituting into the given equation, f'(x) = (1/x) · x² + ln x · 2x Let’s simplify terms to get the final answer, f'(x) = x + 2x ln x Thus, the derivative of the specified function is x + 2x ln 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.
XYZ Company models its revenue with the function y = ln(x), where y represents the revenue at a time x. If x = 2 hours, calculate the rate of change of revenue.
We have y = ln(x) (rate of change of revenue)...(1) Now, we will differentiate equation (1) Take the derivative ln(x): dy/dx = 1/x Given x = 2 (substitute this into the derivative) dy/dx = 1/2 Hence, we get the rate of change of revenue at x = 2 hours as 1/2.
We find the rate of change of the revenue at x = 2 hours as 1/2, which means that at this point, the revenue increases at half the rate per unit time.
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. We then differentiate the first derivative to find the second derivative. This gives us the rate at which the first derivative changes.
Prove: d/dx (ln²(x)) = 2 ln(x)/x.
Let’s start using the chain rule: Consider y = ln²(x) = [ln(x)]² To differentiate, we use the chain rule: dy/dx = 2 ln(x) · d/dx [ln(x)] Since the derivative of ln(x) is 1/x, dy/dx = 2 ln(x) · 1/x Substituting y = ln²(x), d/dx (ln²(x)) = 2 ln(x)/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 substitute y = ln²(x) 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 will 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. Natural Logarithm: The natural logarithm is a logarithm to the base e, where e is approximately equal to 2.71828. Quotient Rule: A rule used for differentiating functions that are the ratio of two differentiable functions. Chain Rule: A rule used for differentiating compositions of functions. Logarithmic Differentiation: A technique used to differentiate functions by taking the logarithm of both sides.
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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