Last updated on July 18th, 2025
We use the derivative of ln(x), which is 1/x, to understand how the natural logarithm function changes in response to a slight change in x. Derivatives are crucial in various applications, including calculating growth rates and decay in natural processes. We will now discuss 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:
Logarithm Function: ln(x) is the natural logarithm of x.
Chain Rule: A rule used to differentiate composite functions involving ln(x).
Quotient Rule: Used when ln(x) appears in a fraction.
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 where x > 0.
We can derive the derivative of ln(x) using proofs. To show this, we will use the definition of the natural logarithm and the rules of differentiation. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of ln(x) results in 1/x using the above-mentioned methods:
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, we will 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 = limₕ→₀ [ln((x + h)/x)] / h = limₕ→₀ [1/x] * [ln(1 + h/x) / (h/x)]
As h approaches 0, ln(1 + h/x) / (h/x) approaches 1 (using the limit property of ln), f'(x) = 1/x Hence, proved.
To prove the differentiation of ln(x) using the chain rule, Consider that we have a composite function involving ln(x), such as ln(u(x)), where u(x) is a differentiable function. The chain rule states that: d/dx [ln(u(x))] = (1/u(x)) * u'(x) For the simplest case where u(x) = x, we have: d/dx [ln(x)] = (1/x) * 1 = 1/x
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can provide deeper insights into the behavior of functions. For example, the second derivative can tell us about the concavity of the function.
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.
When x is 0, the derivative is undefined because ln(x) is undefined at x = 0. When x is 1, the derivative of ln(x) = 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(x² + 1).
Here, we have f(x) = ln(x² + 1). Using the chain rule, f'(x) = 1/(x² + 1) * d/dx(x² + 1) = 1/(x² + 1) * 2x = 2x/(x² + 1) Thus, the derivative of the specified function is 2x/(x² + 1).
We find the derivative of the given function using the chain rule. The first step is finding the derivative of the inner function and then combining it using the chain rule to get the final result.
A population grows according to the function P(t) = ln(t + 1), where t is time in years. Calculate the rate of change of the population at t = 4 years.
We have P(t) = ln(t + 1) (population function)...(1)
Now, we will differentiate the equation (1) Take the derivative: dP/dt = 1/(t + 1)
Substitute t = 4 into the derivative: dP/dt = 1/(4 + 1) = 1/5
Hence, the rate of change of the population at t = 4 years is 1/5 or 0.2.
We find the rate of change of the population at t = 4 years as 1/5, which indicates the rate at which the population is increasing at that specific 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 1/x to find the second derivative, simplifying 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) = ln(u), where u = x²
To differentiate, we use the chain rule: dy/dx = 1/u * du/dx = 1/(x²) * 2x = 2/x Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace the inner function 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 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.
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