Derivative of ln(-5x)
2026-02-28 06:11 Diff
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Last updated on August 5, 2025

We use the derivative of ln(-5x) as a tool for understanding how logarithmic functions change in response to a slight change in x. Derivatives help us calculate various real-life scenarios, including growth rates and decay. We will now discuss the derivative of ln(-5x) in detail.

What is the Derivative of ln(-5x)?

We now explore the derivative of ln(-5x). It is commonly represented as d/dx (ln(-5x)) or (ln(-5x))', and its value is -1/x. The function ln(-5x) has a clearly defined derivative, indicating it is differentiable within its applicable domain. The key concepts are mentioned below: Logarithmic Function: ln(-5x) represents the natural logarithm of -5x. Chain Rule: Used for differentiating composite functions like ln(-5x). Reciprocal Rule: Involved in differentiating ln(x) to obtain 1/x as the derivative.

Derivative of ln(-5x) Formula

The derivative of ln(-5x) can be denoted as d/dx (ln(-5x)) or (ln(-5x))'. The formula we use for differentiation is: d/dx (ln(-5x)) = -1/x The formula applies to all x where -5x is positive, that is, when x < 0.

Proofs of the Derivative of ln(-5x)

We can derive the derivative of ln(-5x) 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: Using Chain Rule To prove the differentiation of ln(-5x) using the chain rule: Consider f(x) = ln(u) where u = -5x. By the chain rule: d/dx [ln(u)] = (1/u)·(du/dx) Let u = -5x, so du/dx = -5. Substituting these, we get: d/dx (ln(-5x)) = (1/(-5x))·(-5) = -1/x Hence, proved.

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Higher-Order Derivatives of ln(-5x)

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 velocity (first derivative) and acceleration (second derivative) for a moving object. Higher-order derivatives help us understand the behavior of functions like ln(-5x). The first derivative of a function is denoted as f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative and is denoted using f′′(x). Similarly, the third derivative, f′′′(x), results from the second derivative, and this pattern continues. For the nth derivative of ln(-5x), 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.

Special Cases:

When x is 0, the derivative is undefined because ln(-5x) is not defined for non-negative values of x. When x = -1, the derivative of ln(-5x) = -1/(-1) = 1.

Common Mistakes and How to Avoid Them in Derivatives of ln(-5x)

Students frequently make mistakes when differentiating ln(-5x). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:

Problem 1

Calculate the derivative of ln(-5x²).

Okay, lets begin

Here, we have f(x) = ln(-5x²). Using the chain rule, f'(x) = d/dx [ln(-5x²)] Let u = -5x², then du/dx = -10x. f'(x) = (1/u)·(du/dx) = (1/(-5x²))·(-10x) f'(x) = -10x/(-5x²) = 2/x Thus, the derivative of ln(-5x²) is 2/x.

Explanation

We differentiate the given function using the chain rule. By identifying the inner function and its derivative, we apply the chain rule to find the overall derivative.

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Problem 2

A company monitors the growth of a bacteria population with the expression y = ln(-5x), where x represents time in days. Find the rate of change of the population at x = -2 days.

Okay, lets begin

We have y = ln(-5x) (rate of change of population)...(1) Now, we will differentiate equation (1): dy/dx = -1/x Given x = -2 (substitute this into the derivative): dy/dx = -1/(-2) = 1/2 Hence, the rate of change of the population at x = -2 days is 1/2.

Explanation

We find the rate of change of the population at x = -2 by substituting into the derivative of ln(-5x). This gives us the rate at which the population changes with respect to time.

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Problem 3

Derive the second derivative of the function y = ln(-5x).

Okay, lets begin

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(-5x) is 1/x².

Explanation

We start with the first derivative and differentiate it to find the second derivative. This process involves the power rule and results in a positive reciprocal squared.

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Problem 4

Prove: d/dx (ln(-5x³)) = -3/x.

Okay, lets begin

Let’s start using the chain rule: Consider y = ln(-5x³) Let u = -5x³, so du/dx = -15x². By the chain rule: dy/dx = (1/u)·(du/dx) = (1/(-5x³))·(-15x²) = -15x²/(-5x³) = 3/x Hence, d/dx (ln(-5x³)) = 3/x.

Explanation

In this step-by-step process, we use the chain rule to differentiate ln(-5x³). By substituting the derivative of the inner function, we simplify to reach the final result.

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Problem 5

Solve: d/dx (ln(-5x)/x).

Okay, lets begin

To differentiate the function, we use the quotient rule: d/dx (ln(-5x)/x) = (d/dx (ln(-5x))·x - ln(-5x)·d/dx(x))/x² We will substitute d/dx (ln(-5x)) = -1/x and d/dx (x) = 1: ((-1/x)·x - ln(-5x)·1)/x² = (-1 - ln(-5x))/x² Therefore, d/dx (ln(-5x)/x) = (-1 - ln(-5x))/x².

Explanation

We differentiate the given function using the quotient rule. By applying the rule and simplifying, we obtain the final result.

Well explained 👍

FAQs on the Derivative of ln(-5x)

1.Find the derivative of ln(-5x).

Using the chain rule for ln(-5x), we get: d/dx (ln(-5x)) = -1/x.

2.Can we use the derivative of ln(-5x) in real life?

Yes, the derivative of ln(-5x) can be used in real-life scenarios, such as calculating rates of decay or growth in various scientific fields.

3.Is it possible to take the derivative of ln(-5x) at the point where x = 0?

No, ln(-5x) is not defined at x = 0, making it impossible to take the derivative at this point.

4.What rule is used to differentiate ln(-5x)/x?

We use the quotient rule to differentiate ln(-5x)/x: d/dx (ln(-5x)/x) = (x·(-1/x) - ln(-5x)·1)/x².

5.Are the derivatives of ln(x) and ln(-5x) the same?

No, they are different. The derivative of ln(x) is 1/x, whereas the derivative of ln(-5x) is -1/x.

Important Glossaries for the Derivative of ln(-5x)

Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Logarithmic Function: A function that involves the logarithm, such as ln(-5x). Chain Rule: A rule used to differentiate composite functions. Quotient Rule: A rule used for differentiating functions expressed as a quotient of two functions. Domain: The set of input values (x) for which a function is defined.

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Jaskaran Singh Saluja

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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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