1 added
2 removed
Original
2026-01-01
Modified
2026-02-28
1
-
<p>148 Learners</p>
1
+
<p>165 Learners</p>
2
<p>Last updated on<strong>September 12, 2025</strong></p>
2
<p>Last updated on<strong>September 12, 2025</strong></p>
3
<p>We use the derivative of 10/x, which is -10/x², as a tool to measure how the function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of 10/x in detail.</p>
3
<p>We use the derivative of 10/x, which is -10/x², as a tool to measure how the function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of 10/x in detail.</p>
4
<h2>What is the Derivative of 10/x?</h2>
4
<h2>What is the Derivative of 10/x?</h2>
5
<p>We now understand the derivative<a>of</a>10/x. It is commonly represented as d/dx (10/x) or (10/x)', and its value is -10/x².</p>
5
<p>We now understand the derivative<a>of</a>10/x. It is commonly represented as d/dx (10/x) or (10/x)', and its value is -10/x².</p>
6
<p>The<a>function</a>10/x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
6
<p>The<a>function</a>10/x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
7
<p>Reciprocal Function: (10/x).</p>
7
<p>Reciprocal Function: (10/x).</p>
8
<p>Power Rule: Rule for differentiating x raised to a<a>power</a>.</p>
8
<p>Power Rule: Rule for differentiating x raised to a<a>power</a>.</p>
9
<p>Negative Exponents: Understanding how to differentiate functions with<a>negative exponents</a>.</p>
9
<p>Negative Exponents: Understanding how to differentiate functions with<a>negative exponents</a>.</p>
10
<h2>Derivative of 10/x Formula</h2>
10
<h2>Derivative of 10/x Formula</h2>
11
<p>The derivative of 10/x can be denoted as d/dx (10/x) or (10/x)'.</p>
11
<p>The derivative of 10/x can be denoted as d/dx (10/x) or (10/x)'.</p>
12
<p>The<a>formula</a>we use to differentiate 10/x is: d/dx (10/x) = -10/x² (or) (10/x)' = -10/x²</p>
12
<p>The<a>formula</a>we use to differentiate 10/x is: d/dx (10/x) = -10/x² (or) (10/x)' = -10/x²</p>
13
<p>The formula applies to all x where x ≠ 0.</p>
13
<p>The formula applies to all x where x ≠ 0.</p>
14
<h2>Proofs of the Derivative of 10/x</h2>
14
<h2>Proofs of the Derivative of 10/x</h2>
15
<p>We can derive the derivative of 10/x using proofs. To show this, we will use basic differentiation rules. There are several methods we use to prove this, such as:</p>
15
<p>We can derive the derivative of 10/x using proofs. To show this, we will use basic differentiation rules. 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 Power Rule</li>
17
<li>Using Power 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 10/x results in -10/x² using the above-mentioned methods:</p>
19
</ol><p>We will now demonstrate that the differentiation of 10/x results in -10/x² using the above-mentioned methods:</p>
20
<h3>By First Principle</h3>
20
<h3>By First Principle</h3>
21
<p>The derivative of 10/x can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
21
<p>The derivative of 10/x can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
22
<p>To find the derivative of 10/x using the first principle, we will consider f(x) = 10/x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
22
<p>To find the derivative of 10/x using the first principle, we will consider f(x) = 10/x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
23
<p>Given that f(x) = 10/x, we write f(x + h) = 10/(x + h).</p>
23
<p>Given that f(x) = 10/x, we write f(x + h) = 10/(x + h).</p>
24
<p>Substituting these into<a>equation</a>(1),</p>
24
<p>Substituting these into<a>equation</a>(1),</p>
25
<p>f'(x) = limₕ→₀ [10/(x + h) - 10/x] / h = limₕ→₀ [10x - 10(x + h)] / [hx(x + h)] = limₕ→₀ [-10h] / [hx(x + h)] = limₕ→₀ [-10] / [x(x + h)] = -10/x²</p>
25
<p>f'(x) = limₕ→₀ [10/(x + h) - 10/x] / h = limₕ→₀ [10x - 10(x + h)] / [hx(x + h)] = limₕ→₀ [-10h] / [hx(x + h)] = limₕ→₀ [-10] / [x(x + h)] = -10/x²</p>
26
<p>Hence, proved.</p>
26
<p>Hence, proved.</p>
27
<h3>Using Power Rule</h3>
27
<h3>Using Power Rule</h3>
28
<p>To prove the differentiation of 10/x using the power rule, Rewrite 10/x as 10x⁻¹.</p>
28
<p>To prove the differentiation of 10/x using the power rule, Rewrite 10/x as 10x⁻¹.</p>
29
<p>Using the power rule: d/dx (xⁿ) = nxⁿ⁻¹ d/dx (10x⁻¹) = 10(-1)x⁻² = -10x⁻²</p>
29
<p>Using the power rule: d/dx (xⁿ) = nxⁿ⁻¹ d/dx (10x⁻¹) = 10(-1)x⁻² = -10x⁻²</p>
30
<p>This simplifies to -10/x². Using Quotient Rule We will now prove the derivative of 10/x using the quotient rule. The step-by-step process is demonstrated below: Let u = 10 and v = x.</p>
30
<p>This simplifies to -10/x². Using Quotient Rule We will now prove the derivative of 10/x using the quotient rule. The step-by-step process is demonstrated below: Let u = 10 and v = x.</p>
31
<p>By quotient rule: d/dx [u/v] = [v u' - u v'] / [v²]</p>
31
<p>By quotient rule: d/dx [u/v] = [v u' - u v'] / [v²]</p>
32
<p>Let’s substitute u = 10 and v = x, d/dx (10/x) = [x(0) - 10(1)] / x² = -10/x² Thus, d/dx (10/x) = -10/x².</p>
32
<p>Let’s substitute u = 10 and v = x, d/dx (10/x) = [x(0) - 10(1)] / x² = -10/x² Thus, d/dx (10/x) = -10/x².</p>
33
<h3>Explore Our Programs</h3>
33
<h3>Explore Our Programs</h3>
34
-
<p>No Courses Available</p>
35
<h2>Higher-Order Derivatives of 10/x</h2>
34
<h2>Higher-Order Derivatives of 10/x</h2>
36
<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. 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 10/x.</p>
35
<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. 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 10/x.</p>
37
<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>
36
<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>
38
<p>For the nth Derivative of 10/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>
37
<p>For the nth Derivative of 10/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>
39
<h2>Special Cases:</h2>
38
<h2>Special Cases:</h2>
40
<p>When x is 0, the derivative is undefined because 10/x has a discontinuity there. When x is 1, the derivative of 10/x = -10/1², which is -10.</p>
39
<p>When x is 0, the derivative is undefined because 10/x has a discontinuity there. When x is 1, the derivative of 10/x = -10/1², which is -10.</p>
41
<h2>Common Mistakes and How to Avoid Them in Derivatives of 10/x</h2>
40
<h2>Common Mistakes and How to Avoid Them in Derivatives of 10/x</h2>
42
<p>Students frequently make mistakes when differentiating 10/x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
41
<p>Students frequently make mistakes when differentiating 10/x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
43
<h3>Problem 1</h3>
42
<h3>Problem 1</h3>
44
<p>Calculate the derivative of (10/x)·x²</p>
43
<p>Calculate the derivative of (10/x)·x²</p>
45
<p>Okay, lets begin</p>
44
<p>Okay, lets begin</p>
46
<p>Here, we have f(x) = (10/x)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 10/x and v = x².</p>
45
<p>Here, we have f(x) = (10/x)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 10/x and v = x².</p>
47
<p>Let’s differentiate each term, u′= d/dx (10/x) = -10/x² v′= d/dx (x²) = 2x</p>
46
<p>Let’s differentiate each term, u′= d/dx (10/x) = -10/x² v′= d/dx (x²) = 2x</p>
48
<p>Substituting into the given equation, f'(x) = (-10/x²)·x² + (10/x)·2x</p>
47
<p>Substituting into the given equation, f'(x) = (-10/x²)·x² + (10/x)·2x</p>
49
<p>Let’s simplify terms to get the final answer, f'(x) = -10 + 20/x</p>
48
<p>Let’s simplify terms to get the final answer, f'(x) = -10 + 20/x</p>
50
<p>Thus, the derivative of the specified function is -10 + 20/x.</p>
49
<p>Thus, the derivative of the specified function is -10 + 20/x.</p>
51
<h3>Explanation</h3>
50
<h3>Explanation</h3>
52
<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>
51
<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>
53
<p>Well explained 👍</p>
52
<p>Well explained 👍</p>
54
<h3>Problem 2</h3>
53
<h3>Problem 2</h3>
55
<p>A company monitors their production efficiency using the function y = 10/x, where y represents the efficiency at production level x. If x = 5 units, measure the rate of change of efficiency.</p>
54
<p>A company monitors their production efficiency using the function y = 10/x, where y represents the efficiency at production level x. If x = 5 units, measure the rate of change of efficiency.</p>
56
<p>Okay, lets begin</p>
55
<p>Okay, lets begin</p>
57
<p>We have y = 10/x (efficiency function)...(1)</p>
56
<p>We have y = 10/x (efficiency function)...(1)</p>
58
<p>Now, we will differentiate the equation (1)</p>
57
<p>Now, we will differentiate the equation (1)</p>
59
<p>Take the derivative of 10/x: dy/dx = -10/x²</p>
58
<p>Take the derivative of 10/x: dy/dx = -10/x²</p>
60
<p>Given x = 5 (substitute this into the derivative)</p>
59
<p>Given x = 5 (substitute this into the derivative)</p>
61
<p>dy/dx = -10/5² = -10/25 = -2/5</p>
60
<p>dy/dx = -10/5² = -10/25 = -2/5</p>
62
<p>Hence, we get the rate of change of efficiency at a production level of x = 5 as -2/5.</p>
61
<p>Hence, we get the rate of change of efficiency at a production level of x = 5 as -2/5.</p>
63
<h3>Explanation</h3>
62
<h3>Explanation</h3>
64
<p>We find the rate of change of efficiency at x = 5 as -2/5, which means that at this production level, the efficiency decreases as the production level increases.</p>
63
<p>We find the rate of change of efficiency at x = 5 as -2/5, which means that at this production level, the efficiency decreases as the production level increases.</p>
65
<p>Well explained 👍</p>
64
<p>Well explained 👍</p>
66
<h3>Problem 3</h3>
65
<h3>Problem 3</h3>
67
<p>Derive the second derivative of the function y = 10/x.</p>
66
<p>Derive the second derivative of the function y = 10/x.</p>
68
<p>Okay, lets begin</p>
67
<p>Okay, lets begin</p>
69
<p>The first step is to find the first derivative, dy/dx = -10/x²...(1)</p>
68
<p>The first step is to find the first derivative, dy/dx = -10/x²...(1)</p>
70
<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-10/x²] = 20/x³</p>
69
<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-10/x²] = 20/x³</p>
71
<p>Therefore, the second derivative of the function y = 10/x is 20/x³.</p>
70
<p>Therefore, the second derivative of the function y = 10/x is 20/x³.</p>
72
<h3>Explanation</h3>
71
<h3>Explanation</h3>
73
<p>We use the step-by-step process, starting with the first derivative. We then apply the power rule again to find the second derivative, which results in 20/x³.</p>
72
<p>We use the step-by-step process, starting with the first derivative. We then apply the power rule again to find the second derivative, which results in 20/x³.</p>
74
<p>Well explained 👍</p>
73
<p>Well explained 👍</p>
75
<h3>Problem 4</h3>
74
<h3>Problem 4</h3>
76
<p>Prove: d/dx [(10/x)²] = -20/x³</p>
75
<p>Prove: d/dx [(10/x)²] = -20/x³</p>
77
<p>Okay, lets begin</p>
76
<p>Okay, lets begin</p>
78
<p>Let’s start using the chain rule: Consider y = (10/x)² = [10x⁻¹]²</p>
77
<p>Let’s start using the chain rule: Consider y = (10/x)² = [10x⁻¹]²</p>
79
<p>To differentiate, we use the chain rule: dy/dx = 2[10x⁻¹]·d/dx [10x⁻¹]</p>
78
<p>To differentiate, we use the chain rule: dy/dx = 2[10x⁻¹]·d/dx [10x⁻¹]</p>
80
<p>Since the derivative of 10x⁻¹ is -10x⁻², dy/dx = 2[10x⁻¹]·[-10x⁻²] = -20/x³</p>
79
<p>Since the derivative of 10x⁻¹ is -10x⁻², dy/dx = 2[10x⁻¹]·[-10x⁻²] = -20/x³</p>
81
<p>Hence proved.</p>
80
<p>Hence proved.</p>
82
<h3>Explanation</h3>
81
<h3>Explanation</h3>
83
<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace 10x⁻¹ with its derivative. As a final step, we substitute back to derive the equation.</p>
82
<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace 10x⁻¹ with its derivative. As a final step, we substitute back to derive the equation.</p>
84
<p>Well explained 👍</p>
83
<p>Well explained 👍</p>
85
<h3>Problem 5</h3>
84
<h3>Problem 5</h3>
86
<p>Solve: d/dx (10x/x)</p>
85
<p>Solve: d/dx (10x/x)</p>
87
<p>Okay, lets begin</p>
86
<p>Okay, lets begin</p>
88
<p>To differentiate the function, we use the quotient rule: d/dx (10x/x) = (d/dx (10x)·x - 10x·d/dx(x))/x²</p>
87
<p>To differentiate the function, we use the quotient rule: d/dx (10x/x) = (d/dx (10x)·x - 10x·d/dx(x))/x²</p>
89
<p>We will substitute d/dx (10x) = 10 and d/dx (x) = 1 = (10·x - 10x·1)/x² = (10x - 10x)/x² = 0/x²</p>
88
<p>We will substitute d/dx (10x) = 10 and d/dx (x) = 1 = (10·x - 10x·1)/x² = (10x - 10x)/x² = 0/x²</p>
90
<p>Therefore, d/dx (10x/x) = 0</p>
89
<p>Therefore, d/dx (10x/x) = 0</p>
91
<h3>Explanation</h3>
90
<h3>Explanation</h3>
92
<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 of 0.</p>
91
<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 of 0.</p>
93
<p>Well explained 👍</p>
92
<p>Well explained 👍</p>
94
<h2>FAQs on the Derivative of 10/x</h2>
93
<h2>FAQs on the Derivative of 10/x</h2>
95
<h3>1.Find the derivative of 10/x.</h3>
94
<h3>1.Find the derivative of 10/x.</h3>
96
<p>Using the power rule for 10x⁻¹, d/dx (10/x) = -10/x² (simplified).</p>
95
<p>Using the power rule for 10x⁻¹, d/dx (10/x) = -10/x² (simplified).</p>
97
<h3>2.Can we use the derivative of 10/x in real life?</h3>
96
<h3>2.Can we use the derivative of 10/x in real life?</h3>
98
<p>Yes, we can use the derivative of 10/x in real life to analyze rates of change in various fields such as physics, economics, and engineering.</p>
97
<p>Yes, we can use the derivative of 10/x in real life to analyze rates of change in various fields such as physics, economics, and engineering.</p>
99
<h3>3.Is it possible to take the derivative of 10/x at the point where x = 0?</h3>
98
<h3>3.Is it possible to take the derivative of 10/x at the point where x = 0?</h3>
100
<p>No, x = 0 is a point where 10/x is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
99
<p>No, x = 0 is a point where 10/x is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
101
<h3>4.What rule is used to differentiate 10/x?</h3>
100
<h3>4.What rule is used to differentiate 10/x?</h3>
102
<p>We use the power rule or the quotient rule to differentiate 10/x, depending on how the function is presented.</p>
101
<p>We use the power rule or the quotient rule to differentiate 10/x, depending on how the function is presented.</p>
103
<h3>5.Are the derivatives of 10/x and x/10 the same?</h3>
102
<h3>5.Are the derivatives of 10/x and x/10 the same?</h3>
104
<p>No, they are different. The derivative of 10/x is -10/x², while the derivative of x/10 is 1/10.</p>
103
<p>No, they are different. The derivative of 10/x is -10/x², while the derivative of x/10 is 1/10.</p>
105
<h2>Important Glossaries for the Derivative of 10/x</h2>
104
<h2>Important Glossaries for the Derivative of 10/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>Reciprocal Function:</strong>A function in the form of a constant divided by a variable, such as 10/x.</li>
106
</ul><ul><li><strong>Reciprocal Function:</strong>A function in the form of a constant divided by a variable, such as 10/x.</li>
108
</ul><ul><li><strong>Power Rule:</strong>A basic rule in differentiation used to find the derivative of x raised to any power.</li>
107
</ul><ul><li><strong>Power Rule:</strong>A basic rule in differentiation used to find the derivative of x raised to any power.</li>
109
</ul><ul><li><strong>Quotient Rule:</strong>A rule in differentiation used to find the derivative of the quotient of two functions.</li>
108
</ul><ul><li><strong>Quotient Rule:</strong>A rule in differentiation used to find the derivative of the quotient of two functions.</li>
110
</ul><ul><li><strong>Undefined:</strong>Refers to the points where a function does not have a value, often leading to discontinuities.</li>
109
</ul><ul><li><strong>Undefined:</strong>Refers to the points where a function does not have a value, often leading to discontinuities.</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>