Last updated on July 22nd, 2025
We use the derivative of ln(1+1/x), which is -1/(x(x+1)), to understand how the logarithmic 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(1+1/x) in detail.
We now understand the derivative of ln(1+1/x). It is commonly represented as d/dx (ln(1+1/x)) or (ln(1+1/x))', and its value is -1/(x(x+1)). The function ln(1+1/x) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Logarithmic Function: ln(1+1/x).
Chain Rule: Rule for differentiating composite functions like ln(1+1/x).
Reciprocal Rule: Used in simplifying the derivative.
The derivative of ln(1+1/x) can be denoted as d/dx (ln(1+1/x)) or (ln(1+1/x))'. The formula we use to differentiate ln(1+1/x) is: d/dx (ln(1+1/x)) = -1/(x(x+1))
The formula applies to all x where x ≠ -1 or 0.
We can derive the derivative of ln(1+1/x) using proofs. To show this, we will use chain and reciprocal rules along with differentiation.
There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of ln(1+1/x) results in -1/(x(x+1)) using the above-mentioned methods:
To prove the differentiation of ln(1+1/x) using the chain rule, Let u = 1+1/x Then ln(1+1/x) = ln(u) By chain rule: d/dx (ln(u)) = (1/u) * (du/dx)
So, we differentiate u with respect to x: du/dx = d/dx (1+1/x) = -1/x²
Therefore, d/dx (ln(1+1/x)) = (1/(1+1/x)) * (-1/x²) = -1/(x(x+1))
To prove the differentiation of ln(1+1/x) using the quotient rule, Start with the function u = 1+1/x, then differentiate using the quotient rule:
Let u = 1 and v = x, then u/v = 1/x.
d/dx (1/x) = (v * du/dx - u * dv/dx) / v² = (x * 0 - 1 * 1) / x² = -1/x²
Thus, d/dx (ln(1+1/x)) = 1/(1+1/x) * (-1/x²) = -1/(x(x+1))
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(1+1/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(1+1/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 -1 or 0, the derivative is undefined because the function ln(1+1/x) has a vertical asymptote there. When x is 1, the derivative of ln(1+1/x) = -1/(1(1+1)) = -1/2.
Students frequently make mistakes when differentiating ln(1+1/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(1+1/x)²
Here, we have f(x) = ln(1+1/x)².
Using the chain rule, f'(x) = 2 * ln(1+1/x) * d/dx (ln(1+1/x)) In the given equation, d/dx (ln(1+1/x)) = -1/(x(x+1)).
Substituting this into the equation, f'(x) = 2 * ln(1+1/x) * (-1/(x(x+1)))
Let's simplify terms to get the final answer, f'(x) = -2 ln(1+1/x)/(x(x+1))
Thus, the derivative of the specified function is -2 ln(1+1/x)/(x(x+1)).
We find the derivative of the given function by using the chain rule. The first step is finding its derivative and then simplifying the terms to get the final result.
AXB International School invested in a project where the logarithmic growth of resources is modeled by y = ln(1+1/x), where y represents the growth at time x. If x = 2, find the rate of growth.
We have y = ln(1+1/x) (growth model)...(1)
Now, we will differentiate the equation (1) Take the derivative ln(1+1/x): dy/dx = -1/(x(x+1))
Given x = 2 (substitute this into the derivative)
dy/dx = -1/(2(2+1)) = -1/6
Hence, we get the rate of growth at x=2 as -1/6.
We find the rate of growth at x=2 as -1/6, which means that at this point, the growth rate is decreasing slightly.
Derive the second derivative of the function y = ln(1+1/x).
The first step is to find the first derivative, dy/dx = -1/(x(x+1))...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/(x(x+1))]
Using the quotient rule, d²y/dx² = d/dx [x(x+1)] * (-1) - (-1) * [d/dx (x(x+1))] / [x(x+1)]² = [-(x+1) - x] / [x(x+1)]³ = -2x - 1 / [x³(x+1)³]
Therefore, the second derivative of the function y = ln(1+1/x) is -2x - 1 / [x³(x+1)³].
We use the step-by-step process, where we start with the first derivative. Using the quotient rule, we differentiate the function again. We then simplify the terms to find the final answer.
Prove: d/dx (ln(1+1/x)³) = 3 ln(1+1/x)² * (-1/(x(x+1))).
Let’s start using the chain rule: Consider y = ln(1+1/x)³ = [ln(1+1/x)]³
To differentiate, we use the chain rule: dy/dx = 3 [ln(1+1/x)]² * d/dx [ln(1+1/x)]
Since the derivative of ln(1+1/x) is -1/(x(x+1)), dy/dx = 3 [ln(1+1/x)]² * (-1/(x(x+1)))
Substituting y = ln(1+1/x)³, d/dx (ln(1+1/x)³) = 3 ln(1+1/x)² * (-1/(x(x+1)))
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace ln(1+1/x) with its derivative. As a final step, we substitute y = ln(1+1/x)³ to derive the equation.
Solve: d/dx (ln(1+1/x)/x)
To differentiate the function, we use the quotient rule: d/dx (ln(1+1/x)/x) = (d/dx (ln(1+1/x)) * x - ln(1+1/x) * d/dx(x))/x²
We will substitute d/dx (ln(1+1/x)) = -1/(x(x+1)) and d/dx (x) = 1 = (-1/(x(x+1)) * x - ln(1+1/x) * 1) / x² = (-1/(x+1) - ln(1+1/x)) / x²
Therefore, d/dx (ln(1+1/x)/x) = (-1/(x+1) - ln(1+1/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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