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Last updated on September 26, 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 logarithm to the base e, where e is the Euler's number.
Chain Rule: A rule for differentiating composite functions.
Negative Sign: The derivative introduces a negative sign due to the nature of the function.
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 chain rule along with the rules of differentiation. There are several methods we use to prove this, such as:
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 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/x+h)/(x/x))] / h = limₕ→₀ [ln(1 + h/x)] / h
Using the limit property ln(1 + y) ≈ y for small y, f'(x) = limₕ→₀ [(h/x) / h] = -1/x Hence, proved.
To prove the differentiation of ln(-x) using the chain rule, We use the formula: ln(-x) = ln(u), where u = -x
By the chain rule: d/dx [ln(-x)] = (1/u) * (du/dx)
Let’s substitute u = -x, d/dx (ln(-x)) = (1/(-x)) * (-1) = -1/x Hence, proved.
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 f n(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 there.
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(-3x).
Here, we have f(x) = ln(-3x).
Using the chain rule, f'(x) = (1/(-3x)) * (-3) = -1/x
Thus, the derivative of ln(-3x) is -1/x.
We find the derivative of the given function by applying the chain rule.
The first step is finding its derivative and then simplifying to get the final result.
The temperature T(x) of a chemical reaction is represented by T(x) = ln(-4x), where x is the time in minutes. Find the rate of change of temperature at x = -2 minutes.
We have T(x) = ln(-4x) (temperature function)...(1)
Now, we will differentiate the equation (1).
Take the derivative of ln(-4x): dT/dx = -1/x
Given x = -2 (substitute this into the derivative), dT/dx = -1/(-2) = 1/2
Hence, the rate of change of temperature at x = -2 minutes is 1/2.
We find the rate of change of temperature at x = -2 as 1/2, which means that at that point, the temperature increases at a rate of 0.5 units per minute.
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)(-1/x²) = 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 get the second derivative, simplifying the terms to find the final answer.
Prove: d/dx (ln(-5x²)) = -2/x.
Let’s start using the chain rule: Consider y = ln(-5x²) = ln(-(5x²))
To differentiate, we use the chain rule: dy/dx = (1/(-5x²)) * (-10x) = -2/x Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation.
Then, we replace the 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.
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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