1 added
2 removed
Original
2026-01-01
Modified
2026-02-28
1
-
<p>183 Learners</p>
1
+
<p>204 Learners</p>
2
<p>Last updated on<strong>August 5, 2025</strong></p>
2
<p>Last updated on<strong>August 5, 2025</strong></p>
3
<p>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.</p>
3
<p>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.</p>
4
<h2>What is the Derivative of ln(1+1/x)?</h2>
4
<h2>What is the Derivative of ln(1+1/x)?</h2>
5
<p>We now understand the derivative<a>of</a>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<a>function</a>ln(1+1/x) has a clearly defined derivative, indicating it is differentiable within its domain.</p>
5
<p>We now understand the derivative<a>of</a>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<a>function</a>ln(1+1/x) has a clearly defined derivative, indicating it is differentiable within its domain.</p>
6
<p>The key concepts are mentioned below:</p>
6
<p>The key concepts are mentioned below:</p>
7
<p>Logarithmic Function: ln(1+1/x).</p>
7
<p>Logarithmic Function: ln(1+1/x).</p>
8
<p>Chain Rule: Rule for differentiating composite functions like ln(1+1/x).</p>
8
<p>Chain Rule: Rule for differentiating composite functions like ln(1+1/x).</p>
9
<p>Reciprocal Rule: Used in simplifying the derivative.</p>
9
<p>Reciprocal Rule: Used in simplifying the derivative.</p>
10
<h2>Derivative of ln(1+1/x) Formula</h2>
10
<h2>Derivative of ln(1+1/x) Formula</h2>
11
<p>The derivative of ln(1+1/x) can be denoted as d/dx (ln(1+1/x)) or (ln(1+1/x))'. The<a>formula</a>we use to differentiate ln(1+1/x) is: d/dx (ln(1+1/x)) = -1/(x(x+1))</p>
11
<p>The derivative of ln(1+1/x) can be denoted as d/dx (ln(1+1/x)) or (ln(1+1/x))'. The<a>formula</a>we use to differentiate ln(1+1/x) is: d/dx (ln(1+1/x)) = -1/(x(x+1))</p>
12
<p>The formula applies to all x where x ≠ -1 or 0.</p>
12
<p>The formula applies to all x where x ≠ -1 or 0.</p>
13
<h2>Proofs of the Derivative of ln(1+1/x)</h2>
13
<h2>Proofs of the Derivative of ln(1+1/x)</h2>
14
<p>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.</p>
14
<p>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.</p>
15
<p>There are several methods we use to prove this, such as:</p>
15
<p>There are several methods we use to prove this, such as:</p>
16
<ol><li>By First Principle</li>
16
<ol><li>By First Principle</li>
17
<li>Using Chain Rule</li>
17
<li>Using Chain Rule</li>
18
<li>Using Quotient Rule</li>
18
<li>Using Quotient Rule</li>
19
</ol><p>We will now demonstrate that the differentiation of ln(1+1/x) results in -1/(x(x+1)) using the above-mentioned methods:</p>
19
</ol><p>We will now demonstrate that the differentiation of ln(1+1/x) results in -1/(x(x+1)) using the above-mentioned methods:</p>
20
<h3>Using Chain Rule</h3>
20
<h3>Using Chain Rule</h3>
21
<p>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)</p>
21
<p>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)</p>
22
<p>So, we differentiate u with respect to x: du/dx = d/dx (1+1/x) = -1/x²</p>
22
<p>So, we differentiate u with respect to x: du/dx = d/dx (1+1/x) = -1/x²</p>
23
<p>Therefore, d/dx (ln(1+1/x)) = (1/(1+1/x)) * (-1/x²) = -1/(x(x+1))</p>
23
<p>Therefore, d/dx (ln(1+1/x)) = (1/(1+1/x)) * (-1/x²) = -1/(x(x+1))</p>
24
<h3>Using Quotient Rule</h3>
24
<h3>Using Quotient Rule</h3>
25
<p>To prove the differentiation of ln(1+1/x) using the<a>quotient</a>rule, Start with the function u = 1+1/x, then differentiate using the quotient rule:</p>
25
<p>To prove the differentiation of ln(1+1/x) using the<a>quotient</a>rule, Start with the function u = 1+1/x, then differentiate using the quotient rule:</p>
26
<p>Let u = 1 and v = x, then u/v = 1/x.</p>
26
<p>Let u = 1 and v = x, then u/v = 1/x.</p>
27
<p>d/dx (1/x) = (v * du/dx - u * dv/dx) / v² = (x * 0 - 1 * 1) / x² = -1/x²</p>
27
<p>d/dx (1/x) = (v * du/dx - u * dv/dx) / v² = (x * 0 - 1 * 1) / x² = -1/x²</p>
28
<p>Thus, d/dx (ln(1+1/x)) = 1/(1+1/x) * (-1/x²) = -1/(x(x+1))</p>
28
<p>Thus, d/dx (ln(1+1/x)) = 1/(1+1/x) * (-1/x²) = -1/(x(x+1))</p>
29
<h3>Explore Our Programs</h3>
29
<h3>Explore Our Programs</h3>
30
-
<p>No Courses Available</p>
31
<h2>Higher-Order Derivatives of ln(1+1/x)</h2>
30
<h2>Higher-Order Derivatives of ln(1+1/x)</h2>
32
<p>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.</p>
31
<p>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.</p>
33
<p>To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like ln(1+1/x).</p>
32
<p>To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like ln(1+1/x).</p>
34
<p>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.</p>
33
<p>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.</p>
35
<p>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).</p>
34
<p>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).</p>
36
<h2>Special Cases</h2>
35
<h2>Special Cases</h2>
37
<p>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.</p>
36
<p>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.</p>
38
<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(1+1/x)</h2>
37
<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(1+1/x)</h2>
39
<p>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:</p>
38
<p>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:</p>
40
<h3>Problem 1</h3>
39
<h3>Problem 1</h3>
41
<p>Calculate the derivative of ln(1+1/x)²</p>
40
<p>Calculate the derivative of ln(1+1/x)²</p>
42
<p>Okay, lets begin</p>
41
<p>Okay, lets begin</p>
43
<p>Here, we have f(x) = ln(1+1/x)².</p>
42
<p>Here, we have f(x) = ln(1+1/x)².</p>
44
<p>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)).</p>
43
<p>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)).</p>
45
<p>Substituting this into the equation, f'(x) = 2 * ln(1+1/x) * (-1/(x(x+1)))</p>
44
<p>Substituting this into the equation, f'(x) = 2 * ln(1+1/x) * (-1/(x(x+1)))</p>
46
<p>Let's simplify terms to get the final answer, f'(x) = -2 ln(1+1/x)/(x(x+1))</p>
45
<p>Let's simplify terms to get the final answer, f'(x) = -2 ln(1+1/x)/(x(x+1))</p>
47
<p>Thus, the derivative of the specified function is -2 ln(1+1/x)/(x(x+1)).</p>
46
<p>Thus, the derivative of the specified function is -2 ln(1+1/x)/(x(x+1)).</p>
48
<h3>Explanation</h3>
47
<h3>Explanation</h3>
49
<p>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.</p>
48
<p>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.</p>
50
<p>Well explained 👍</p>
49
<p>Well explained 👍</p>
51
<h3>Problem 2</h3>
50
<h3>Problem 2</h3>
52
<p>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.</p>
51
<p>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.</p>
53
<p>Okay, lets begin</p>
52
<p>Okay, lets begin</p>
54
<p>We have y = ln(1+1/x) (growth model)...(1)</p>
53
<p>We have y = ln(1+1/x) (growth model)...(1)</p>
55
<p>Now, we will differentiate the equation (1) Take the derivative ln(1+1/x): dy/dx = -1/(x(x+1))</p>
54
<p>Now, we will differentiate the equation (1) Take the derivative ln(1+1/x): dy/dx = -1/(x(x+1))</p>
56
<p>Given x = 2 (substitute this into the derivative)</p>
55
<p>Given x = 2 (substitute this into the derivative)</p>
57
<p>dy/dx = -1/(2(2+1)) = -1/6</p>
56
<p>dy/dx = -1/(2(2+1)) = -1/6</p>
58
<p>Hence, we get the rate of growth at x=2 as -1/6.</p>
57
<p>Hence, we get the rate of growth at x=2 as -1/6.</p>
59
<h3>Explanation</h3>
58
<h3>Explanation</h3>
60
<p>We find the rate of growth at x=2 as -1/6, which means that at this point, the growth rate is decreasing slightly.</p>
59
<p>We find the rate of growth at x=2 as -1/6, which means that at this point, the growth rate is decreasing slightly.</p>
61
<p>Well explained 👍</p>
60
<p>Well explained 👍</p>
62
<h3>Problem 3</h3>
61
<h3>Problem 3</h3>
63
<p>Derive the second derivative of the function y = ln(1+1/x).</p>
62
<p>Derive the second derivative of the function y = ln(1+1/x).</p>
64
<p>Okay, lets begin</p>
63
<p>Okay, lets begin</p>
65
<p>The first step is to find the first derivative, dy/dx = -1/(x(x+1))...(1)</p>
64
<p>The first step is to find the first derivative, dy/dx = -1/(x(x+1))...(1)</p>
66
<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/(x(x+1))]</p>
65
<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/(x(x+1))]</p>
67
<p>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]</p>
66
<p>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]</p>
68
<p>Therefore, the second derivative of the function y = ln(1+1/x) is -2x - 1 / [x³(x+1)³].</p>
67
<p>Therefore, the second derivative of the function y = ln(1+1/x) is -2x - 1 / [x³(x+1)³].</p>
69
<h3>Explanation</h3>
68
<h3>Explanation</h3>
70
<p>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.</p>
69
<p>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.</p>
71
<p>Well explained 👍</p>
70
<p>Well explained 👍</p>
72
<h3>Problem 4</h3>
71
<h3>Problem 4</h3>
73
<p>Prove: d/dx (ln(1+1/x)³) = 3 ln(1+1/x)² * (-1/(x(x+1))).</p>
72
<p>Prove: d/dx (ln(1+1/x)³) = 3 ln(1+1/x)² * (-1/(x(x+1))).</p>
74
<p>Okay, lets begin</p>
73
<p>Okay, lets begin</p>
75
<p>Let’s start using the chain rule: Consider y = ln(1+1/x)³ = [ln(1+1/x)]³</p>
74
<p>Let’s start using the chain rule: Consider y = ln(1+1/x)³ = [ln(1+1/x)]³</p>
76
<p>To differentiate, we use the chain rule: dy/dx = 3 [ln(1+1/x)]² * d/dx [ln(1+1/x)]</p>
75
<p>To differentiate, we use the chain rule: dy/dx = 3 [ln(1+1/x)]² * d/dx [ln(1+1/x)]</p>
77
<p>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)))</p>
76
<p>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)))</p>
78
<p>Substituting y = ln(1+1/x)³, d/dx (ln(1+1/x)³) = 3 ln(1+1/x)² * (-1/(x(x+1)))</p>
77
<p>Substituting y = ln(1+1/x)³, d/dx (ln(1+1/x)³) = 3 ln(1+1/x)² * (-1/(x(x+1)))</p>
79
<p>Hence proved.</p>
78
<p>Hence proved.</p>
80
<h3>Explanation</h3>
79
<h3>Explanation</h3>
81
<p>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.</p>
80
<p>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.</p>
82
<p>Well explained 👍</p>
81
<p>Well explained 👍</p>
83
<h3>Problem 5</h3>
82
<h3>Problem 5</h3>
84
<p>Solve: d/dx (ln(1+1/x)/x)</p>
83
<p>Solve: d/dx (ln(1+1/x)/x)</p>
85
<p>Okay, lets begin</p>
84
<p>Okay, lets begin</p>
86
<p>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</p>
85
<p>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</p>
87
<p>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</p>
86
<p>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</p>
88
<p>Therefore, d/dx (ln(1+1/x)/x) = (-1/(x+1) - ln(1+1/x)) / x²</p>
87
<p>Therefore, d/dx (ln(1+1/x)/x) = (-1/(x+1) - ln(1+1/x)) / x²</p>
89
<h3>Explanation</h3>
88
<h3>Explanation</h3>
90
<p>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.</p>
89
<p>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.</p>
91
<p>Well explained 👍</p>
90
<p>Well explained 👍</p>
92
<h2>FAQs on the Derivative of ln(1+1/x)</h2>
91
<h2>FAQs on the Derivative of ln(1+1/x)</h2>
93
<h3>1.Find the derivative of ln(1+1/x).</h3>
92
<h3>1.Find the derivative of ln(1+1/x).</h3>
94
<p>Using the chain rule with ln(1+1/x), d/dx (ln(1+1/x)) = -1/(x(x+1)) (simplified)</p>
93
<p>Using the chain rule with ln(1+1/x), d/dx (ln(1+1/x)) = -1/(x(x+1)) (simplified)</p>
95
<h3>2.Can we use the derivative of ln(1+1/x) in real life?</h3>
94
<h3>2.Can we use the derivative of ln(1+1/x) in real life?</h3>
96
<p>Yes, we can use the derivative of ln(1+1/x) in real life in calculating the rate of change of any logarithmic growth, especially in fields such as mathematics, physics, and economics.</p>
95
<p>Yes, we can use the derivative of ln(1+1/x) in real life in calculating the rate of change of any logarithmic growth, especially in fields such as mathematics, physics, and economics.</p>
97
<h3>3.Is it possible to take the derivative of ln(1+1/x) at the point where x = -1?</h3>
96
<h3>3.Is it possible to take the derivative of ln(1+1/x) at the point where x = -1?</h3>
98
<p>No, x = -1 is a point where ln(1+1/x) is undefined, so it is impossible to take the derivative at these points (since the function does not exist there).</p>
97
<p>No, x = -1 is a point where ln(1+1/x) is undefined, so it is impossible to take the derivative at these points (since the function does not exist there).</p>
99
<h3>4.What rule is used to differentiate ln(1+1/x)/x?</h3>
98
<h3>4.What rule is used to differentiate ln(1+1/x)/x?</h3>
100
<p>We use the quotient rule to differentiate ln(1+1/x)/x, d/dx (ln(1+1/x)/x) = (-1/(x+1) - ln(1+1/x)) / x².</p>
99
<p>We use the quotient rule to differentiate ln(1+1/x)/x, d/dx (ln(1+1/x)/x) = (-1/(x+1) - ln(1+1/x)) / x².</p>
101
<h3>5.Are the derivatives of ln(1+1/x) and ln(x) the same?</h3>
100
<h3>5.Are the derivatives of ln(1+1/x) and ln(x) the same?</h3>
102
<p>No, they are different. The derivative of ln(1+1/x) is -1/(x(x+1)), while the derivative of ln(x) is 1/x.</p>
101
<p>No, they are different. The derivative of ln(1+1/x) is -1/(x(x+1)), while the derivative of ln(x) is 1/x.</p>
103
<h3>6.Can we find the derivative of the ln(1+1/x) formula?</h3>
102
<h3>6.Can we find the derivative of the ln(1+1/x) formula?</h3>
104
<p>To find, consider y = ln(1+1/x). We use the chain rule: y’ = (1/(1+1/x)) * (-1/x²) = -1/(x(x+1)).</p>
103
<p>To find, consider y = ln(1+1/x). We use the chain rule: y’ = (1/(1+1/x)) * (-1/x²) = -1/(x(x+1)).</p>
105
<h2>Important Glossaries for the Derivative of ln(1+1/x)</h2>
104
<h2>Important Glossaries for the Derivative of ln(1+1/x)</h2>
106
<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
105
<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
107
</ul><ul><li><strong>Logarithmic Function:</strong>A function that involves the logarithm, such as ln(1+1/x).</li>
106
</ul><ul><li><strong>Logarithmic Function:</strong>A function that involves the logarithm, such as ln(1+1/x).</li>
108
</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating composite functions by differentiating the outer and inner functions.</li>
107
</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating composite functions by differentiating the outer and inner functions.</li>
109
</ul><ul><li><strong>Quotient Rule:</strong>A rule used for finding the derivative of a quotient of two functions.</li>
108
</ul><ul><li><strong>Quotient Rule:</strong>A rule used for finding the derivative of a quotient of two functions.</li>
110
</ul><ul><li><strong>Undefined:</strong>A term describing a point where a function does not have a defined value, often leading to division by zero.</li>
109
</ul><ul><li><strong>Undefined:</strong>A term describing a point where a function does not have a defined value, often leading to division by zero.</li>
111
</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
110
</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
112
<p>▶</p>
111
<p>▶</p>
113
<h2>Jaskaran Singh Saluja</h2>
112
<h2>Jaskaran Singh Saluja</h2>
114
<h3>About the Author</h3>
113
<h3>About the Author</h3>
115
<p>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.</p>
114
<p>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.</p>
116
<h3>Fun Fact</h3>
115
<h3>Fun Fact</h3>
117
<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
116
<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>