Last updated on July 15th, 2025
We use the derivative of 4ln(x), which is 4/x, as a tool to understand how the natural logarithmic function changes with respect to x. Derivatives are essential in real-life applications like calculating growth rates or decay. We will now discuss the derivative of 4ln(x) in detail.
We will explore the derivative of 4ln(x). It is commonly represented as d/dx (4lnx) or (4lnx)', and its value is 4/x. The function 4ln(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 logarithmic function. Constant Multiplication Rule: This rule helps differentiate a constant times a function. Reciprocal Function: 1/x is the derivative of ln(x).
The derivative of 4ln(x) can be denoted as d/dx (4lnx) or (4lnx)'. The formula we use to differentiate 4ln(x) is: d/dx (4lnx) = 4/x (or) (4lnx)' = 4/x The formula applies to all x > 0.
We can derive the derivative of 4ln(x) using proofs. To show this, we will use the rules of differentiation. There are several methods we use to prove this, such as: Using Constant Multiplication Rule Using the Chain Rule We will now demonstrate that the differentiation of 4ln(x) results in 4/x using the above-mentioned methods: Using Constant Multiplication Rule The derivative of 4ln(x) can be proved using the constant multiplication rule. d/dx [4ln(x)] = 4 * d/dx [ln(x)] Since d/dx [ln(x)] = 1/x, d/dx [4ln(x)] = 4 * (1/x) = 4/x Hence, proved. Using the Chain Rule To prove the differentiation of 4ln(x) using the chain rule, Consider f(x) = 4ln(x) and g(x) = x We differentiate f(g(x)): d/dx [4ln(x)] = 4 * d/dx [ln(x)] As d/dx [ln(x)] = 1/x, d/dx [4ln(x)] = 4 * (1/x) = 4/x Thus, the derivative is proved.
When a function is differentiated multiple times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, consider a scenario where you are tracking the growth rate (first derivative) and the acceleration of that growth rate (second derivative). Higher-order derivatives make it easier to understand functions like 4ln(x). For the first derivative of a function, we write f′(x), indicating 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 4ln(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.
When x is 0, the derivative is undefined because ln(x) is undefined at x = 0. When x is 1, the derivative of 4ln(x) = 4/1, which is 4.
Students frequently make mistakes when differentiating 4ln(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 (4lnx · x²)
Here, we have f(x) = 4lnx · x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 4lnx and v = x². Let’s differentiate each term, u′= d/dx (4lnx) = 4/x v′= d/dx (x²) = 2x Substituting into the given equation, f'(x) = (4/x) · x² + (4lnx) · 2x Let’s simplify terms to get the final answer, f'(x) = 4x + 8xlnx Thus, the derivative of the specified function is 4x + 8xlnx.
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.
The growth rate of a population is modeled by the function y = 4ln(x), where y represents the growth at a time x. If x = 5 years, measure the growth rate.
We have y = 4ln(x) (growth rate of the population)...(1) Now, we will differentiate the equation (1) Take the derivative 4ln(x): dy/dx = 4/x Given x = 5 (substitute this into the derivative) dy/dx = 4/5 Hence, the growth rate at x = 5 years is 4/5.
We find the growth rate at x = 5 years as 4/5, which indicates that at this point, the growth rate is 0.8 times the unit time period.
Derive the second derivative of the function y = 4ln(x).
The first step is to find the first derivative, dy/dx = 4/x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [4/x] d²y/dx² = -4/x² Therefore, the second derivative of the function y = 4ln(x) is -4/x².
We use the step-by-step process, where we start with the first derivative. We then differentiate 4/x to find the second derivative, resulting in -4/x².
Prove: d/dx (4ln²(x)) = 8ln(x)/x
Let’s start using the chain rule: Consider y = 4ln²(x) = [4(ln(x))²] To differentiate, we use the chain rule: dy/dx = 4 * 2ln(x) * d/dx [ln(x)] Since the derivative of ln(x) is 1/x, dy/dx = 8ln(x)/x Substituting y = 4ln²(x), d/dx (4ln²(x)) = 8ln(x)/x Hence proved.
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 y = 4ln²(x) to derive the equation.
Solve: d/dx (4lnx/x)
To differentiate the function, we use the quotient rule: d/dx (4lnx/x) = (d/dx (4lnx) · x - 4lnx · d/dx(x))/x² We will substitute d/dx (4lnx) = 4/x and d/dx (x) = 1 = (4/x · x - 4lnx · 1) / x² = (4 - 4lnx) / x² Therefore, d/dx (4lnx/x) = (4 - 4lnx) / 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.
Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Natural Logarithm: The logarithm to the base e, represented as ln(x). Constant Multiplication Rule: A rule stating that the derivative of a constant times a function is the constant times the derivative of the function. Quotient Rule: A rule used to differentiate functions that are the quotient of two other functions. Chain Rule: A rule for differentiating compositions of functions.
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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