Last updated on July 18th, 2025
We use the derivative of ln(x/3), which is 1/x, as a tool to understand 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/3) in detail.
We now understand the derivative of ln(x/3). It is commonly represented as d/dx [ln(x/3)] or [ln(x/3)]', and its value is 1/x. The function ln(x/3) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Natural Logarithm: ln(x/3) is the logarithm of x/3 with base e. Chain Rule: Rule for differentiating composite functions like ln(x/3). Reciprocal Rule: Used for finding the derivative of functions like 1/x.
The derivative of ln(x/3) can be denoted as d/dx [ln(x/3)] or [ln(x/3)]'. The formula we use to differentiate ln(x/3) is: d/dx [ln(x/3)] = 1/x The formula applies to all x where x > 0.
We can derive the derivative of ln(x/3) using proofs. To show this, we will use the properties of logarithms along with the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using Chain Rule Using Properties of Logarithms We will now demonstrate that the differentiation of ln(x/3) results in 1/x using the above-mentioned methods: By First Principle The derivative of ln(x/3) 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/3) using the first principle, consider f(x) = ln(x/3). 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/3), we write f(x + h) = ln((x + h)/3). Substituting these into the equation, f'(x) = limₕ→₀ [ln((x + h)/3) - ln(x/3)] / h = limₕ→₀ [ln((x + h)/x)] / h = limₕ→₀ [ln(1 + h/x)] / h Using the property 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/3) using the chain rule, We use the formula: ln(x/3) = ln(x) - ln(3) The derivative of ln(x) is 1/x, and the derivative of a constant ln(3) is 0. Thus, d/dx [ln(x/3)] = d/dx [ln(x) - ln(3)] = 1/x - 0 = 1/x Using Properties of Logarithms We will now prove the derivative of ln(x/3) using the properties of logarithms. Here is the step-by-step process: ln(x/3) = ln(x) - ln(3) The derivative of ln(x) is 1/x, and the derivative of a constant ln(3) is 0. d/dx [ln(x/3)] = d/dx [ln(x) - ln(3)] = 1/x - 0 = 1/x Thus, the derivative of ln(x/3) 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/3). 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/3), 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 = 0, the derivative is undefined because ln(x/3) is not defined for non-positive values of x. When x = 3, the derivative of ln(x/3) = 1/3, which is the reciprocal of the input value.
Students frequently make mistakes when differentiating ln(x/3). 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/3)·x²
Here, we have f(x) = ln(x/3)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln(x/3) and v = x². Let’s differentiate each term, u′ = d/dx [ln(x/3)] = 1/x v′ = d/dx [x²] = 2x Substituting into the given equation, f'(x) = (1/x)·x² + ln(x/3)·2x Let’s simplify terms to get the final answer, f'(x) = x + 2x ln(x/3) Thus, the derivative of the specified function is x + 2x ln(x/3).
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.
A company is studying the growth rate of its customer base. The growth is modeled by the function y = ln(x/3), where y represents the growth and x represents time in months. If x = 6 months, find the rate of growth.
We have y = ln(x/3) (growth function)...(1) Now, we will differentiate the equation (1) Take the derivative ln(x/3): dy/dx = 1/x Given x = 6 (substitute this into the derivative) dy/dx = 1/6 Hence, we get the rate of growth at x = 6 months as 1/6.
We find the rate of growth at x = 6 months as 1/6, which means that at this point, the growth rate is 1/6 per month.
Derive the second derivative of the function y = ln(x/3).
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] d²y/dx² = -1/x² Therefore, the second derivative of the function y = ln(x/3) is -1/x².
We use the step-by-step process, starting with the first derivative. By differentiating 1/x, we find the second derivative is -1/x².
Prove: d/dx [ln((x/3)²)] = 2/x.
Let’s start using the chain rule: Consider y = ln((x/3)²) = 2 ln(x/3) To differentiate, we use the chain rule: dy/dx = 2 d/dx [ln(x/3)] Since the derivative of ln(x/3) is 1/x, dy/dx = 2(1/x) Substituting y = ln((x/3)²), d/dx [ln((x/3)²)] = 2/x Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace ln(x/3) with its derivative. As a final step, we substitute y = ln((x/3)²) to derive the equation.
Solve: d/dx (ln(x/3)/x)
To differentiate the function, we use the quotient rule: d/dx [ln(x/3)/x] = (d/dx [ln(x/3)]·x - ln(x/3)·d/dx(x)) / x² We will substitute d/dx [ln(x/3)] = 1/x and d/dx(x) = 1 = (1/x·x - ln(x/3)·1) / x² = (1 - ln(x/3)) / x² Therefore, d/dx [ln(x/3)/x] = (1 - ln(x/3)) / x²
In this process, we differentiate the given function using the product rule and 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: A logarithmic function with base e, usually written as ln(x). Chain Rule: A rule used to differentiate composite functions. Quotient Rule: A method for differentiating the ratio of two functions. Logarithmic Properties: Rules that simplify the differentiation of logarithmic expressions.
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