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Last updated on July 21st, 2025

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Derivative of ln Squared

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We use the derivative of (ln(x))² as a tool to understand the rate of change of the natural logarithm squared function with respect to x. Derivatives are crucial in various fields such as physics, engineering, and economics for understanding growth, decay, and optimization. We will now discuss the derivative of (ln(x))² in detail.

Derivative of ln Squared for UAE Students
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What is the Derivative of ln Squared?

We now explore the derivative of (ln(x))². It is commonly represented as d/dx ((ln(x))²) or ((ln(x))²)', and its value is 2ln(x)/x. This function has a clearly defined derivative, indicating it is differentiable within its domain (x > 0). The key concepts involved are mentioned below: Natural Logarithm Function: ln(x) is the natural logarithm of x. Chain Rule: Rule for differentiating composite functions like (ln(x))².

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Derivative of ln Squared Formula

The derivative of (ln(x))² can be denoted as d/dx ((ln(x))²) or ((ln(x))²)'. The formula we use to differentiate (ln(x))² is: d/dx ((ln(x))²) = 2ln(x)/x This formula applies to all x > 0.

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Proofs of the Derivative of ln Squared

We can derive the derivative of (ln(x))² using different methods. To show this, we will use the chain rule along with the properties of logarithms. Below are a few methods to prove this: Using Chain Rule To prove the differentiation of (ln(x))² using the chain rule, we consider it as a composition of functions: Let u = ln(x), so u² = (ln(x))². Using the chain rule: d/dx (u²) = 2u · du/dx Substituting u = ln(x) and du/dx = 1/x, we have: d/dx ((ln(x))²) = 2ln(x) · (1/x) = 2ln(x)/x Using Product Rule We can alternatively consider (ln(x))² as ln(x) · ln(x) and use the product rule: d/dx (ln(x) · ln(x)) = ln(x) · (1/x) + ln(x) · (1/x) = 2ln(x)/x Thus, we obtain the same result.

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Higher-Order Derivatives of ln Squared

When a function is differentiated multiple times, the derivatives obtained are called higher-order derivatives. Understanding higher-order derivatives can be a bit complex, but they provide insights into the curvature and concavity of functions. For example, the first derivative gives us the slope, while the second derivative tells us about the rate of change of the slope. For the first derivative of a function, we write f′(x), indicating the function's change or its slope at a certain point. The second derivative, denoted as f′′(x), is derived from the first derivative and indicates the concavity of the function. For the nth derivative of (ln(x))², we generally use fⁿ(x) to represent the nth derivative, which describes the change in the rate of change.

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Special Cases:

At x = 1, the derivative of (ln(x))² simplifies to 2ln(1)/1, which is 0, since ln(1) = 0. When x approaches 0 from the positive side, the derivative becomes undefined because ln(x) is not defined for non-positive values.

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Common Mistakes and How to Avoid Them in Derivatives of ln Squared

Students frequently make mistakes when differentiating (ln(x))². These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:

Mistake 1

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Not Applying the Chain Rule

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Students often forget to use the chain rule. This occurs when the derivative of the inner function is not considered. For example: Incorrect: d/dx ((ln(x))²) = 2ln(x). To fix this error, always identify the inner function (ln(x)) and apply the chain rule properly, which includes finding the derivative of ln(x) as 1/x.

Mistake 2

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Incorrect Simplification

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Another common mistake involves incorrect simplification of the derivative expression. Students may simplify d/dx ((ln(x))²) = 2/x ln(x) instead of 2ln(x)/x. Ensure each step is simplified correctly, preserving the structure of the expression.

Mistake 3

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Forgetting the Domain of ln(x)

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Students might not remember that ln(x) is only defined for x > 0. Always consider the domain of the function for correct differentiation, especially when applying chain or product rules.

Mistake 4

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Misapplying Product Rule

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When considering (ln(x))² as ln(x) · ln(x), students might misapply the product rule. For example: Incorrect: d/dx (ln(x) · ln(x)) = ln(x)/x. Correct application of the product rule gives 2ln(x)/x.

Mistake 5

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Not Considering Higher-Order Derivatives

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Students often overlook the importance of higher-order derivatives, such as the second derivative, which provides information on concavity. Make sure to calculate higher-order derivatives for a deeper understanding of the function's behavior.

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Examples Using the Derivative of ln Squared

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Problem 1

Calculate the derivative of (ln(x)·x³).

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Here, we have f(x) = ln(x) · x³. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln(x) and v = x³. Let's differentiate each term, u′ = d/dx (ln(x)) = 1/x v′ = d/dx (x³) = 3x² Substituting into the given equation, f'(x) = (1/x) · x³ + ln(x) · 3x² Let's simplify terms to get the final answer, f'(x) = x² + 3x²ln(x) Thus, the derivative of the specified function is x² + 3x²ln(x).

Explanation

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.

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Problem 2

A research company measures the growth rate of bacteria, represented by the function y = (ln(x))², where y represents the growth rate at a time x. If x = e, measure the growth rate.

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We have y = (ln(x))² (growth rate formula)...(1) Now, we will differentiate equation (1): Take the derivative of (ln(x))²: dy/dx = 2ln(x)/x Given x = e (substitute this into the derivative), dy/dx = 2ln(e)/e Since ln(e) = 1, dy/dx = 2/e Hence, we get the growth rate of the bacteria at x = e as 2/e.

Explanation

We find the growth rate of the bacteria at x = e as 2/e, indicating that at this point, the rate of change of growth is 2/e times the initial rate.

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Problem 3

Derive the second derivative of the function y = (ln(x))².

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The first step is to find the first derivative, dy/dx = 2ln(x)/x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2ln(x)/x] Here we use the quotient rule, d²y/dx² = (x · (2/x) - 2ln(x) · 1)/(x²) = (2 - 2ln(x))/(x²) Therefore, the second derivative of the function y = (ln(x))² is (2 - 2ln(x))/(x²).

Explanation

We use the step-by-step process, starting with the first derivative. Using the quotient rule, we differentiate 2ln(x)/x, then simplify the expression to find the final answer.

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Problem 4

Prove: d/dx ((ln(x))³) = 3(ln(x))²/x.

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Let’s start using the chain rule: Consider y = (ln(x))³ = [ln(x)]³ To differentiate, we use the chain rule: dy/dx = 3[ln(x)]² · d/dx [ln(x)] Since the derivative of ln(x) is 1/x, dy/dx = 3[ln(x)]² · (1/x) Substituting y = (ln(x))³, d/dx ((ln(x))³) = 3(ln(x))²/x Hence proved.

Explanation

In this step-by-step process, we use the chain rule to differentiate the equation. Then, we replace the derivative of ln(x) and simplify the expression to prove the equation.

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Problem 5

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

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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

Explanation

In this process, we differentiate the given function using the quotient rule. We substitute the derivatives of each term and simplify the equation to obtain the final result.

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FAQs on the Derivative of ln Squared

1.Find the derivative of (ln(x))².

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

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3.Is it possible to take the derivative of (ln(x))² at the point where x = 0?

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

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5.Are the derivatives of (ln(x))² and (ln(x))³ the same?

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Important Glossaries for the Derivative of ln Squared

Derivative: The derivative of a function measures how the function changes as its input changes. Natural Logarithm: ln(x) is the logarithm to the base e, where e is an irrational constant approximately equal to 2.718. Chain Rule: A rule for differentiating compositions of functions. Quotient Rule: A rule used for differentiating the quotient of two functions. Higher-order Derivative: Derivatives of a function taken multiple times, providing information on the curvature and concavity of the function.

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Jaskaran Singh Saluja

About the Author

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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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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