1 added
2 removed
Original
2026-01-01
Modified
2026-02-28
1
-
<p>198 Learners</p>
1
+
<p>237 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 sin(1/x) as a tool to understand how the sine function changes in response to a slight change in x. Derivatives help us calculate variations in a wide range of real-life situations. We will now discuss the derivative of sin(1/x) in detail.</p>
3
<p>We use the derivative of sin(1/x) as a tool to understand how the sine function changes in response to a slight change in x. Derivatives help us calculate variations in a wide range of real-life situations. We will now discuss the derivative of sin(1/x) in detail.</p>
4
<h2>What is the Derivative of Sin(1/x)?</h2>
4
<h2>What is the Derivative of Sin(1/x)?</h2>
5
<p>We now understand the derivative of sin(1/x). It is commonly represented as d/dx [sin(1/x)] or [sin(1/x)]', and its value involves the chain rule and is given by cos(1/x) * (-1/x²). The<a>function</a>sin(1/x) has a clearly defined derivative, indicating it is differentiable within its domain.</p>
5
<p>We now understand the derivative of sin(1/x). It is commonly represented as d/dx [sin(1/x)] or [sin(1/x)]', and its value involves the chain rule and is given by cos(1/x) * (-1/x²). The<a>function</a>sin(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>Sine Function: sin(θ) is a trigonometric function.</p>
7
<p>Sine Function: sin(θ) is a trigonometric function.</p>
8
<p>Chain Rule: A rule for differentiating compositions<a>of functions</a>, essential for sin(1/x).</p>
8
<p>Chain Rule: A rule for differentiating compositions<a>of functions</a>, essential for sin(1/x).</p>
9
<p>Cosine Function: cos(θ) is another basic trigonometric function.</p>
9
<p>Cosine Function: cos(θ) is another basic trigonometric function.</p>
10
<h2>Derivative of Sin(1/x) Formula</h2>
10
<h2>Derivative of Sin(1/x) Formula</h2>
11
<p>The derivative of sin(1/x) can be denoted as d/dx [sin(1/x)] or [sin(1/x)]'. The<a>formula</a>we use to differentiate sin(1/x) is: d/dx [sin(1/x)] = cos(1/x) * (-1/x²) The formula applies to all x where x ≠ 0, as the function is undefined at x = 0.</p>
11
<p>The derivative of sin(1/x) can be denoted as d/dx [sin(1/x)] or [sin(1/x)]'. The<a>formula</a>we use to differentiate sin(1/x) is: d/dx [sin(1/x)] = cos(1/x) * (-1/x²) The formula applies to all x where x ≠ 0, as the function is undefined at x = 0.</p>
12
<h2>Proofs of the Derivative of Sin(1/x)</h2>
12
<h2>Proofs of the Derivative of Sin(1/x)</h2>
13
<p>We can derive the derivative of sin(1/x) using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
13
<p>We can derive the derivative of sin(1/x) using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
14
<ol><li>By First Principle</li>
14
<ol><li>By First Principle</li>
15
<li>Using Chain Rule</li>
15
<li>Using Chain Rule</li>
16
</ol><p>We will now demonstrate that the differentiation of sin(1/x) results in cos(1/x) * (-1/x²) using the above-mentioned methods:</p>
16
</ol><p>We will now demonstrate that the differentiation of sin(1/x) results in cos(1/x) * (-1/x²) using the above-mentioned methods:</p>
17
<h3>Using Chain Rule</h3>
17
<h3>Using Chain Rule</h3>
18
<p>To prove the differentiation of sin(1/x) using the chain rule, Consider y = sin(1/x). Let u = 1/x, then y = sin(u).</p>
18
<p>To prove the differentiation of sin(1/x) using the chain rule, Consider y = sin(1/x). Let u = 1/x, then y = sin(u).</p>
19
<p>By the chain rule: dy/dx = dy/du * du/dx dy/du = cos(u) = cos(1/x) du/dx = -1/x² Thus, dy/dx = cos(1/x) * (-1/x²)</p>
19
<p>By the chain rule: dy/dx = dy/du * du/dx dy/du = cos(u) = cos(1/x) du/dx = -1/x² Thus, dy/dx = cos(1/x) * (-1/x²)</p>
20
<p>Hence, proved.</p>
20
<p>Hence, proved.</p>
21
<h3>Explore Our Programs</h3>
21
<h3>Explore Our Programs</h3>
22
-
<p>No Courses Available</p>
23
<h2>Higher-Order Derivatives of Sin(1/x)</h2>
22
<h2>Higher-Order Derivatives of Sin(1/x)</h2>
24
<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>
23
<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>
25
<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 help us understand functions like sin(1/x).</p>
24
<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 help us understand functions like sin(1/x).</p>
26
<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>
25
<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>
27
<p>For the nth Derivative of sin(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>
26
<p>For the nth Derivative of sin(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>
28
<h2>Special Cases:</h2>
27
<h2>Special Cases:</h2>
29
<p>When x approaches 0, the function sin(1/x) oscillates and does not have a well-defined limit, leading to discontinuity. When x is large, the derivative cos(1/x) * (-1/x²) approaches 0, indicating the function becomes flatter.</p>
28
<p>When x approaches 0, the function sin(1/x) oscillates and does not have a well-defined limit, leading to discontinuity. When x is large, the derivative cos(1/x) * (-1/x²) approaches 0, indicating the function becomes flatter.</p>
30
<h2>Common Mistakes and How to Avoid Them in Derivatives of Sin(1/x)</h2>
29
<h2>Common Mistakes and How to Avoid Them in Derivatives of Sin(1/x)</h2>
31
<p>Students frequently make mistakes when differentiating sin(1/x). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
30
<p>Students frequently make mistakes when differentiating sin(1/x). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
32
<h3>Problem 1</h3>
31
<h3>Problem 1</h3>
33
<p>Calculate the derivative of (sin(1/x) * e^(1/x))</p>
32
<p>Calculate the derivative of (sin(1/x) * e^(1/x))</p>
34
<p>Okay, lets begin</p>
33
<p>Okay, lets begin</p>
35
<p>Here, we have f(x) = sin(1/x) * e^(1/x).</p>
34
<p>Here, we have f(x) = sin(1/x) * e^(1/x).</p>
36
<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin(1/x) and v = e^(1/x).</p>
35
<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin(1/x) and v = e^(1/x).</p>
37
<p>Let’s differentiate each term, u′ = d/dx [sin(1/x)] = cos(1/x) * (-1/x²) v′ = d/dx [e^(1/x)] = e^(1/x) * (-1/x²)</p>
36
<p>Let’s differentiate each term, u′ = d/dx [sin(1/x)] = cos(1/x) * (-1/x²) v′ = d/dx [e^(1/x)] = e^(1/x) * (-1/x²)</p>
38
<p>Substituting into the given equation, f'(x) = [cos(1/x) * (-1/x²)] * e^(1/x) + sin(1/x) * [e^(1/x) * (-1/x²)]</p>
37
<p>Substituting into the given equation, f'(x) = [cos(1/x) * (-1/x²)] * e^(1/x) + sin(1/x) * [e^(1/x) * (-1/x²)]</p>
39
<p>Let’s simplify terms to get the final answer, f'(x) = (-1/x²) * [cos(1/x) * e^(1/x) + sin(1/x) * e^(1/x)]</p>
38
<p>Let’s simplify terms to get the final answer, f'(x) = (-1/x²) * [cos(1/x) * e^(1/x) + sin(1/x) * e^(1/x)]</p>
40
<p>Thus, the derivative of the specified function is (-1/x²) * e^(1/x) * [cos(1/x) + sin(1/x)].</p>
39
<p>Thus, the derivative of the specified function is (-1/x²) * e^(1/x) * [cos(1/x) + sin(1/x)].</p>
41
<h3>Explanation</h3>
40
<h3>Explanation</h3>
42
<p>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.</p>
41
<p>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.</p>
43
<p>Well explained 👍</p>
42
<p>Well explained 👍</p>
44
<h3>Problem 2</h3>
43
<h3>Problem 2</h3>
45
<p>A company monitors the amplitude of a signal given by y = sin(1/x). If the monitoring point is at x = 0.5, calculate the rate of change of the amplitude.</p>
44
<p>A company monitors the amplitude of a signal given by y = sin(1/x). If the monitoring point is at x = 0.5, calculate the rate of change of the amplitude.</p>
46
<p>Okay, lets begin</p>
45
<p>Okay, lets begin</p>
47
<p>We have y = sin(1/x) (amplitude of the signal)...(1)</p>
46
<p>We have y = sin(1/x) (amplitude of the signal)...(1)</p>
48
<p>Now, we will differentiate the equation (1)</p>
47
<p>Now, we will differentiate the equation (1)</p>
49
<p>Take the derivative of sin(1/x): dy/dx = cos(1/x) * (-1/x²)</p>
48
<p>Take the derivative of sin(1/x): dy/dx = cos(1/x) * (-1/x²)</p>
50
<p>Given x = 0.5 (substitute this into the derivative)</p>
49
<p>Given x = 0.5 (substitute this into the derivative)</p>
51
<p>dy/dx = cos(2) * (-1/(0.5)²) dy/dx = -4 * cos(2)</p>
50
<p>dy/dx = cos(2) * (-1/(0.5)²) dy/dx = -4 * cos(2)</p>
52
<p>Hence, the rate of change of the amplitude at x = 0.5 is -4 * cos(2).</p>
51
<p>Hence, the rate of change of the amplitude at x = 0.5 is -4 * cos(2).</p>
53
<h3>Explanation</h3>
52
<h3>Explanation</h3>
54
<p>We find the rate of change of the amplitude at x = 0.5 by first differentiating the function and then substituting the given value into the derivative.</p>
53
<p>We find the rate of change of the amplitude at x = 0.5 by first differentiating the function and then substituting the given value into the derivative.</p>
55
<p>Well explained 👍</p>
54
<p>Well explained 👍</p>
56
<h3>Problem 3</h3>
55
<h3>Problem 3</h3>
57
<p>Derive the second derivative of the function y = sin(1/x).</p>
56
<p>Derive the second derivative of the function y = sin(1/x).</p>
58
<p>Okay, lets begin</p>
57
<p>Okay, lets begin</p>
59
<p>The first step is to find the first derivative, dy/dx = cos(1/x) * (-1/x²)...(1)</p>
58
<p>The first step is to find the first derivative, dy/dx = cos(1/x) * (-1/x²)...(1)</p>
60
<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [cos(1/x) * (-1/x²)]</p>
59
<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [cos(1/x) * (-1/x²)]</p>
61
<p>Using the product rule, d²y/dx² = (-1/x²) * d/dx [cos(1/x)] + cos(1/x) * d/dx [-1/x²] = (-1/x²) * [-sin(1/x) * (-1/x²)] + cos(1/x) * [2/x³] = (1/x⁴) * sin(1/x) + (2/x³) * cos(1/x)</p>
60
<p>Using the product rule, d²y/dx² = (-1/x²) * d/dx [cos(1/x)] + cos(1/x) * d/dx [-1/x²] = (-1/x²) * [-sin(1/x) * (-1/x²)] + cos(1/x) * [2/x³] = (1/x⁴) * sin(1/x) + (2/x³) * cos(1/x)</p>
62
<p>Therefore, the second derivative of the function y = sin(1/x) is (1/x⁴) * sin(1/x) + (2/x³) * cos(1/x).</p>
61
<p>Therefore, the second derivative of the function y = sin(1/x) is (1/x⁴) * sin(1/x) + (2/x³) * cos(1/x).</p>
63
<h3>Explanation</h3>
62
<h3>Explanation</h3>
64
<p>We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate each component. We then substitute the identities and simplify the terms to find the final answer.</p>
63
<p>We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate each component. We then substitute the identities and simplify the terms to find the final answer.</p>
65
<p>Well explained 👍</p>
64
<p>Well explained 👍</p>
66
<h3>Problem 4</h3>
65
<h3>Problem 4</h3>
67
<p>Prove: d/dx [sin²(1/x)] = 2 sin(1/x) * cos(1/x) * (-1/x²).</p>
66
<p>Prove: d/dx [sin²(1/x)] = 2 sin(1/x) * cos(1/x) * (-1/x²).</p>
68
<p>Okay, lets begin</p>
67
<p>Okay, lets begin</p>
69
<p>Let’s start using the chain rule: Consider y = sin²(1/x) = [sin(1/x)]²</p>
68
<p>Let’s start using the chain rule: Consider y = sin²(1/x) = [sin(1/x)]²</p>
70
<p>To differentiate, we use the chain rule: dy/dx = 2 sin(1/x) * d/dx [sin(1/x)]</p>
69
<p>To differentiate, we use the chain rule: dy/dx = 2 sin(1/x) * d/dx [sin(1/x)]</p>
71
<p>Since the derivative of sin(1/x) is cos(1/x) * (-1/x²), dy/dx = 2 sin(1/x) * [cos(1/x) * (-1/x²)]</p>
70
<p>Since the derivative of sin(1/x) is cos(1/x) * (-1/x²), dy/dx = 2 sin(1/x) * [cos(1/x) * (-1/x²)]</p>
72
<p>Substituting y = sin²(1/x), d/dx [sin²(1/x)] = 2 sin(1/x) * cos(1/x) * (-1/x²)</p>
71
<p>Substituting y = sin²(1/x), d/dx [sin²(1/x)] = 2 sin(1/x) * cos(1/x) * (-1/x²)</p>
73
<p>Hence proved.</p>
72
<p>Hence proved.</p>
74
<h3>Explanation</h3>
73
<h3>Explanation</h3>
75
<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sin(1/x) with its derivative. As a final step, we substitute y = sin²(1/x) to derive the equation.</p>
74
<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sin(1/x) with its derivative. As a final step, we substitute y = sin²(1/x) to derive the equation.</p>
76
<p>Well explained 👍</p>
75
<p>Well explained 👍</p>
77
<h3>Problem 5</h3>
76
<h3>Problem 5</h3>
78
<p>Solve: d/dx [sin(1/x)/x]</p>
77
<p>Solve: d/dx [sin(1/x)/x]</p>
79
<p>Okay, lets begin</p>
78
<p>Okay, lets begin</p>
80
<p>To differentiate the function, we use the quotient rule: d/dx [sin(1/x)/x] = (d/dx [sin(1/x)] * x - sin(1/x) * d/dx[x]) / x²</p>
79
<p>To differentiate the function, we use the quotient rule: d/dx [sin(1/x)/x] = (d/dx [sin(1/x)] * x - sin(1/x) * d/dx[x]) / x²</p>
81
<p>We will substitute d/dx [sin(1/x)] = cos(1/x) * (-1/x²) and d/dx [x] = 1 = [cos(1/x) * (-1/x²) * x - sin(1/x)] / x² = [-cos(1/x) / x³ - sin(1/x)] / x² = [-cos(1/x)/x³ - sin(1/x)/x²]</p>
80
<p>We will substitute d/dx [sin(1/x)] = cos(1/x) * (-1/x²) and d/dx [x] = 1 = [cos(1/x) * (-1/x²) * x - sin(1/x)] / x² = [-cos(1/x) / x³ - sin(1/x)] / x² = [-cos(1/x)/x³ - sin(1/x)/x²]</p>
82
<p>Therefore, d/dx [sin(1/x)/x] = -cos(1/x)/x³ - sin(1/x)/x²</p>
81
<p>Therefore, d/dx [sin(1/x)/x] = -cos(1/x)/x³ - sin(1/x)/x²</p>
83
<h3>Explanation</h3>
82
<h3>Explanation</h3>
84
<p>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.</p>
83
<p>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.</p>
85
<p>Well explained 👍</p>
84
<p>Well explained 👍</p>
86
<h2>FAQs on the Derivative of Sin(1/x)</h2>
85
<h2>FAQs on the Derivative of Sin(1/x)</h2>
87
<h3>1.Find the derivative of sin(1/x).</h3>
86
<h3>1.Find the derivative of sin(1/x).</h3>
88
<p>Using the chain rule, sin(1/x) gives cos(1/x) * (-1/x²).</p>
87
<p>Using the chain rule, sin(1/x) gives cos(1/x) * (-1/x²).</p>
89
<h3>2.Can we use the derivative of sin(1/x) in real life?</h3>
88
<h3>2.Can we use the derivative of sin(1/x) in real life?</h3>
90
<p>Yes, we can use the derivative of sin(1/x) in real life to analyze rates of change, especially in fields such as engineering and physics.</p>
89
<p>Yes, we can use the derivative of sin(1/x) in real life to analyze rates of change, especially in fields such as engineering and physics.</p>
91
<h3>3.Is it possible to take the derivative of sin(1/x) at the point where x = 0?</h3>
90
<h3>3.Is it possible to take the derivative of sin(1/x) at the point where x = 0?</h3>
92
<p>No, x = 0 is a point where sin(1/x) is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
91
<p>No, x = 0 is a point where sin(1/x) is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
93
<h3>4.What rule is used to differentiate sin(1/x)/x?</h3>
92
<h3>4.What rule is used to differentiate sin(1/x)/x?</h3>
94
<p>We use the<a>quotient</a>rule to differentiate sin(1/x)/x, d/dx [sin(1/x)/x] = [cos(1/x) * (-1/x²) * x - sin(1/x)] / x².</p>
93
<p>We use the<a>quotient</a>rule to differentiate sin(1/x)/x, d/dx [sin(1/x)/x] = [cos(1/x) * (-1/x²) * x - sin(1/x)] / x².</p>
95
<h3>5.Are the derivatives of sin(1/x) and sin⁻¹x the same?</h3>
94
<h3>5.Are the derivatives of sin(1/x) and sin⁻¹x the same?</h3>
96
<p>No, they are different. The derivative of sin(1/x) is cos(1/x) * (-1/x²), while the derivative of sin⁻¹x is 1/√(1-x²).</p>
95
<p>No, they are different. The derivative of sin(1/x) is cos(1/x) * (-1/x²), while the derivative of sin⁻¹x is 1/√(1-x²).</p>
97
<h2>Important Glossaries for the Derivative of Sin(1/x)</h2>
96
<h2>Important Glossaries for the Derivative of Sin(1/x)</h2>
98
<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>
97
<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>
99
</ul><ul><li><strong>Chain Rule:</strong>A rule for finding the derivative of a composition of functions.</li>
98
</ul><ul><li><strong>Chain Rule:</strong>A rule for finding the derivative of a composition of functions.</li>
100
</ul><ul><li><strong>Sine Function:</strong>A primary trigonometric function written as sin(θ).</li>
99
</ul><ul><li><strong>Sine Function:</strong>A primary trigonometric function written as sin(θ).</li>
101
</ul><ul><li><strong>Cosine Function:</strong>A trigonometric function represented as cos(θ), the derivative of the sine function.</li>
100
</ul><ul><li><strong>Cosine Function:</strong>A trigonometric function represented as cos(θ), the derivative of the sine function.</li>
102
</ul><ul><li><strong>Quotient Rule:</strong>A technique used to differentiate functions that are divided by each other.</li>
101
</ul><ul><li><strong>Quotient Rule:</strong>A technique used to differentiate functions that are divided by each other.</li>
103
</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
102
</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
104
<p>▶</p>
103
<p>▶</p>
105
<h2>Jaskaran Singh Saluja</h2>
104
<h2>Jaskaran Singh Saluja</h2>
106
<h3>About the Author</h3>
105
<h3>About the Author</h3>
107
<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>
106
<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>
108
<h3>Fun Fact</h3>
107
<h3>Fun Fact</h3>
109
<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
108
<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>