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Last updated on July 15th, 2025

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Derivative of 5lnx

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We use the derivative of 5ln(x), which is 5/x, as a tool for understanding how the logarithmic function changes in response to changes in x. Derivatives are crucial in various fields, including economics and engineering, to calculate rates of change. Let's explore the derivative of 5ln(x) in detail.

Derivative of 5lnx for Canadian Students
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What is the Derivative of 5lnx?

The derivative of 5ln(x) is commonly represented as d/dx (5lnx) or (5lnx)', and its value is 5/x. The function 5ln(x) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Logarithmic Function: ln(x) is the natural logarithm of x. Constant Multiplier Rule: Rule for differentiating functions multiplied by a constant. Derivative of ln(x): The basic derivative of ln(x) is 1/x.

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Derivative of 5lnx Formula

The derivative of 5ln(x) can be denoted as d/dx (5lnx) or (5lnx)'. The formula we use to differentiate 5ln(x) is: d/dx (5lnx) = 5 * d/dx (lnx) = 5/x The formula applies to all x>0.

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Proofs of the Derivative of 5lnx

We can derive the derivative of 5ln(x) using proofs. To show this, we will use the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using the Constant Multiplier Rule By First Principle The derivative of 5ln(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 5ln(x) using the first principle, we consider f(x) = 5ln(x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = 5ln(x), we write f(x + h) = 5ln(x + h). Substituting these into the equation, f'(x) = limₕ→₀ [5ln(x + h) - 5ln(x)] / h = 5 * limₕ→₀ [ln(x + h) - ln(x)] / h = 5 * limₕ→₀ [ln((x + h)/x)] / h = 5 * limₕ→₀ [ln(1 + h/x)] / h = 5 * (1/x) * limₕ→₀ [ln(1 + u)] / u (where u = h/x) Using limit formulas, limₕ→₀ ln(1 + u)/u = 1. f'(x) = 5 * (1/x) = 5/x Hence, proved. Using the Constant Multiplier Rule To prove the differentiation of 5ln(x) using the constant multiplier rule, We use the formula: d/dx (c * f(x)) = c * d/dx (f(x)) Let f(x) = ln(x) and c = 5. The derivative of ln(x) is: d/dx (ln(x)) = 1/x So, d/dx (5ln(x)) = 5 * (1/x) = 5/x

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Higher-Order Derivatives of 5lnx

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's speed (first derivative) and the rate at which the speed changes (second derivative). Higher-order derivatives make it easier to understand functions like 5ln(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 5ln(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 (continuing for higher-order derivatives).

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Special Cases:

When x approaches 0, the derivative becomes undefined because ln(x) is undefined for non-positive values. When x = 1, the derivative of 5ln(x) = 5/1, which is 5.

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Common Mistakes and How to Avoid Them in Derivatives of 5lnx

Students frequently make mistakes when differentiating 5ln(x). 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 constant multiplier

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Students may forget to apply the constant multiplier rule, leading to incorrect results. They often skip this step, especially when solving using basic differentiation rules. Ensure that constants are accounted for in each step. This might seem redundant, but it's crucial to avoid errors.

Mistake 2

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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. Keep in mind the domain of the function you differentiate; it helps understand where the function is continuous.

Mistake 3

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Incorrect Use of Basic Differentiation Rules

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While differentiating functions, students sometimes misapply basic rules. For example: Incorrect differentiation: d/dx (lnx) = x. Correct differentiation: d/dx (lnx) = 1/x. To avoid this mistake, ensure you know the basic derivatives before applying them.

Mistake 4

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Not Writing Constants and Coefficients

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There is a common mistake where students forget to multiply the constants placed before ln(x). For example, they incorrectly write d/dx (5lnx) = 1/x. Students should check the constants in the terms and ensure they are multiplied properly. For example, the correct equation is d/dx (5lnx) = 5/x.

Mistake 5

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Incorrect Application of Limit Concepts

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Students often forget to apply the correct concepts when using limits, especially in first principle derivations. Ensure that limits are applied correctly, with proper substitutions or simplifications.

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Examples Using the Derivative of 5lnx

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Problem 1

Calculate the derivative of (5lnx * x^3).

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Here, we have f(x) = 5lnx * x^3. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 5lnx and v = x^3. Let’s differentiate each term, u′= d/dx (5lnx) = 5/x v′= d/dx (x^3) = 3x^2 Substituting into the given equation, f'(x) = (5/x) * x^3 + 5lnx * 3x^2 Let’s simplify terms to get the final answer, f'(x) = 5x^2 + 15x^2lnx Thus, the derivative of the specified function is 5x^2 + 15x^2lnx.

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 population of bacteria grows exponentially, and its growth rate is modeled by P(t) = 5ln(t) where P represents the population at time t. If t = 2 hours, find the growth rate.

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We have P(t) = 5ln(t) (rate of growth)...(1) Now, we will differentiate the equation (1) Take the derivative of 5ln(t): dP/dt = 5/t Given t = 2 (substitute this into the derivative) dP/dt = 5/2 Hence, we get the growth rate of the population at t = 2 hours as 2.5.

Explanation

We find the growth rate at t = 2 hours as 2.5, which indicates the rate of population increase at that point in time.

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Problem 3

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

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The first step is to find the first derivative, dy/dx = 5/x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [5/x] d²y/dx² = -5/x² Therefore, the second derivative of the function y = 5ln(x) is -5/x².

Explanation

We use the step-by-step process, where we start with the first derivative. We then differentiate again to find the second derivative, simplifying to get the final answer.

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Problem 4

Prove: d/dx ((5lnx)²) = 10ln(x)/x.

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Let’s start using the chain rule: Consider y = (5lnx)² To differentiate, we use the chain rule: dy/dx = 2 * (5lnx) * d/dx (5lnx) Since the derivative of 5ln(x) is 5/x, dy/dx = 2 * (5lnx) * (5/x) = 10ln(x)/x Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace ln(x) with its derivative. As a final step, we substitute and simplify to derive the equation.

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Problem 5

Solve: d/dx (5lnx/x).

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

Explanation

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.

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FAQs on the Derivative of 5lnx

1.Find the derivative of 5lnx.

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2.Can we use the derivative of 5lnx in real life?

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3.Is it possible to take the derivative of 5lnx at x=0?

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4.What rule is used to differentiate 5lnx/x?

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5.Are the derivatives of 5lnx and (lnx)⁵ the same?

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Important Glossaries for the Derivative of 5lnx

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 uses the natural logarithm, represented as ln(x). Constant Multiplier Rule: A rule used in differentiation where a constant multiplied by a function results in the constant multiplied by the derivative of the function. First Derivative: The initial result of differentiating a function, which gives the rate of change of a specific function. Quotient Rule: A rule used to differentiate functions that are divided by each other, expressed as d/dx (u/v) = (v * u' - u * v') / v².

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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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