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2026-01-01
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>We use the derivative of x/5, which is 1/5, to understand how the function changes with respect to a slight change in x. Derivatives can assist us in calculating profit or loss in real-life situations. We will now discuss the derivative of x/5 in detail.</p>
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<p>We use the derivative of x/5, which is 1/5, to understand how the function changes with respect to a slight change in x. Derivatives can assist us in calculating profit or loss in real-life situations. We will now discuss the derivative of x/5 in detail.</p>
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<h2>What is the Derivative of x/5?</h2>
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<h2>What is the Derivative of x/5?</h2>
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<p>We now understand the derivative of x/5. It is commonly represented as d/dx (x/5) or (x/5)', and its value is 1/5.</p>
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<p>We now understand the derivative of x/5. It is commonly represented as d/dx (x/5) or (x/5)', and its value is 1/5.</p>
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<p>The<a>function</a>x/5 has a<a>constant</a>derivative, indicating it is differentiable across its entire domain.</p>
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<p>The<a>function</a>x/5 has a<a>constant</a>derivative, indicating it is differentiable across its entire domain.</p>
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<p>The key concepts include: Linear Function: A function of the form f(x) = mx + b, where m and b are constants.</p>
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<p>The key concepts include: Linear Function: A function of the form f(x) = mx + b, where m and b are constants.</p>
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<p>Constant Rule: The rule for differentiating constant<a>multiples</a><a>of functions</a>.</p>
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<p>Constant Rule: The rule for differentiating constant<a>multiples</a><a>of functions</a>.</p>
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<p>Slope: The measure of steepness of a line, which in this case is the constant 1/5.</p>
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<p>Slope: The measure of steepness of a line, which in this case is the constant 1/5.</p>
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<h2>Derivative of x/5 Formula</h2>
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<h2>Derivative of x/5 Formula</h2>
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<p>The derivative of x/5 can be denoted as d/dx (x/5) or (x/5)'.</p>
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<p>The derivative of x/5 can be denoted as d/dx (x/5) or (x/5)'.</p>
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<p>The<a>formula</a>we use to differentiate x/5 is: d/dx (x/5) = 1/5 (or) (x/5)' = 1/5</p>
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<p>The<a>formula</a>we use to differentiate x/5 is: d/dx (x/5) = 1/5 (or) (x/5)' = 1/5</p>
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<p>The formula applies to all x in the<a>real number</a><a>set</a>.</p>
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<p>The formula applies to all x in the<a>real number</a><a>set</a>.</p>
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<h2>Proofs of the Derivative of x/5</h2>
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<h2>Proofs of the Derivative of x/5</h2>
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<p>We can derive the derivative of x/5 using proofs. To show this, we will use basic differentiation rules.</p>
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<p>We can derive the derivative of x/5 using proofs. To show this, we will use basic differentiation rules.</p>
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<p>There are several methods we use to prove this, such as:</p>
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<p>There are several methods we use to prove this, such as:</p>
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<p>By First Principle Using the Constant Rule We will now demonstrate that the differentiation of x/5 results in 1/5 using these methods:</p>
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<p>By First Principle Using the Constant Rule We will now demonstrate that the differentiation of x/5 results in 1/5 using these methods:</p>
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<p>By First Principle The derivative of x/5 can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>By First Principle The derivative of x/5 can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>To find the derivative of x/5 using the first principle, we consider f(x) = x/5.</p>
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<p>To find the derivative of x/5 using the first principle, we consider f(x) = x/5.</p>
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<p>Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = x/5, we write f(x + h) = (x + h)/5.</p>
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<p>Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = x/5, we write f(x + h) = (x + h)/5.</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [(x + h)/5 - x/5] / h = limₕ→₀ [x/5 + h/5 - x/5] / h = limₕ→₀ [h/5] / h = limₕ→₀ 1/5 Therefore, f'(x) = 1/5.</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [(x + h)/5 - x/5] / h = limₕ→₀ [x/5 + h/5 - x/5] / h = limₕ→₀ [h/5] / h = limₕ→₀ 1/5 Therefore, f'(x) = 1/5.</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p>Using the Constant Rule To prove the differentiation of x/5 using the constant rule, We recognize that x/5 is a linear function with a slope of 1/5. By the constant multiple rule: d/dx (c·f(x)) = c·d/dx (f(x)) Let f(x) = x and c = 1/5, d/dx (x/5) = 1/5·d/dx (x) Since d/dx (x) = 1, d/dx (x/5) = 1/5·1 = 1/5.</p>
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<p>Using the Constant Rule To prove the differentiation of x/5 using the constant rule, We recognize that x/5 is a linear function with a slope of 1/5. By the constant multiple rule: d/dx (c·f(x)) = c·d/dx (f(x)) Let f(x) = x and c = 1/5, d/dx (x/5) = 1/5·d/dx (x) Since d/dx (x) = 1, d/dx (x/5) = 1/5·1 = 1/5.</p>
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<h2>Higher-Order Derivatives of x/5</h2>
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<h2>Higher-Order Derivatives of x/5</h2>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can provide additional insights into the behavior of functions.</p>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can provide additional insights into the behavior of functions.</p>
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<p>For the function x/5, which is linear, the first derivative is 1/5, and all higher-order derivatives are 0.</p>
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<p>For the function x/5, which is linear, the first derivative is 1/5, and all higher-order derivatives are 0.</p>
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<p>This indicates that the function's<a>rate</a>of change is constant, and there are no changes in the rate of change.</p>
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<p>This indicates that the function's<a>rate</a>of change is constant, and there are no changes in the rate of change.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>The derivative of x/5 is constant and defined everywhere on the real line. The derivative does not change with the value of x, making it straightforward to compute.</p>
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<p>The derivative of x/5 is constant and defined everywhere on the real line. The derivative does not change with the value of x, making it straightforward to compute.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of x/5</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of x/5</h2>
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<p>Students frequently make mistakes when differentiating constant multiples of x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
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<p>Students frequently make mistakes when differentiating constant multiples of x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
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<h3>Problem 1</h3>
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<h3>Problem 1</h3>
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<p>Calculate the derivative of (x/5 + 7).</p>
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<p>Calculate the derivative of (x/5 + 7).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Here, we have f(x) = x/5 + 7. The derivative of a constant is 0, so we only differentiate x/5.</p>
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<p>Here, we have f(x) = x/5 + 7. The derivative of a constant is 0, so we only differentiate x/5.</p>
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<p>Using basic rules of differentiation, f'(x) = d/dx (x/5) + d/dx (7) = 1/5 + 0</p>
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<p>Using basic rules of differentiation, f'(x) = d/dx (x/5) + d/dx (7) = 1/5 + 0</p>
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<p>Therefore, the derivative of the specified function is 1/5.</p>
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<p>Therefore, the derivative of the specified function is 1/5.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the derivative of the given function by differentiating each term separately.</p>
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<p>We find the derivative of the given function by differentiating each term separately.</p>
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<p>The derivative of a constant is 0, and the derivative of x/5 is 1/5.</p>
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<p>The derivative of a constant is 0, and the derivative of x/5 is 1/5.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>A company charges a rate represented by the function y = x/5 dollars per hour for a service. If x = 10 hours, calculate the rate of change of the charge with respect to time.</p>
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<p>A company charges a rate represented by the function y = x/5 dollars per hour for a service. If x = 10 hours, calculate the rate of change of the charge with respect to time.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>We have y = x/5 (rate of charge)...(1) Now, we will differentiate the equation (1)</p>
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<p>We have y = x/5 (rate of charge)...(1) Now, we will differentiate the equation (1)</p>
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<p>Take the derivative of x/5: dy/dx = 1/5 This indicates the rate of change is constant at 1/5 dollars per hour.</p>
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<p>Take the derivative of x/5: dy/dx = 1/5 This indicates the rate of change is constant at 1/5 dollars per hour.</p>
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<p>Given x = 10, the rate of change remains 1/5 dollars per hour.</p>
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<p>Given x = 10, the rate of change remains 1/5 dollars per hour.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the rate of change of charge with respect to time as constant, meaning the cost increases steadily at 1/5 dollars per hour regardless of the number of hours.</p>
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<p>We find the rate of change of charge with respect to time as constant, meaning the cost increases steadily at 1/5 dollars per hour regardless of the number of hours.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function y = x/5.</p>
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<p>Derive the second derivative of the function y = x/5.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>The first step is to find the first derivative, dy/dx = 1/5...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = 1/5...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx (1/5) Since 1/5 is a constant, d²y/dx² = 0</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx (1/5) Since 1/5 is a constant, d²y/dx² = 0</p>
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<p>Therefore, the second derivative of the function y = x/5 is 0.</p>
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<p>Therefore, the second derivative of the function y = x/5 is 0.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use the step-by-step process, starting with the first derivative.</p>
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<p>We use the step-by-step process, starting with the first derivative.</p>
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<p>Since the first derivative is a constant, the second derivative is 0, indicating no change in the rate of change.</p>
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<p>Since the first derivative is a constant, the second derivative is 0, indicating no change in the rate of change.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>Prove: d/dx (3x/5) = 3/5.</p>
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<p>Prove: d/dx (3x/5) = 3/5.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Consider y = 3x/5 To differentiate, we use the constant multiple rule: dy/dx = 3·d/dx (x/5) Since d/dx (x/5) = 1/5, dy/dx = 3·1/5 dy/dx = 3/5 Hence proved.</p>
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<p>Consider y = 3x/5 To differentiate, we use the constant multiple rule: dy/dx = 3·d/dx (x/5) Since d/dx (x/5) = 1/5, dy/dx = 3·1/5 dy/dx = 3/5 Hence proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this step-by-step process, we used the constant multiple rule to differentiate the equation. We replaced x/5 with its derivative and computed the result.</p>
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<p>In this step-by-step process, we used the constant multiple rule to differentiate the equation. We replaced x/5 with its derivative and computed the result.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 5</h3>
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<h3>Problem 5</h3>
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<p>Solve: d/dx (x/5 + x²).</p>
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<p>Solve: d/dx (x/5 + x²).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>To differentiate the function, we use basic rules of differentiation: d/dx (x/5 + x²) = d/dx (x/5) + d/dx (x²)</p>
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<p>To differentiate the function, we use basic rules of differentiation: d/dx (x/5 + x²) = d/dx (x/5) + d/dx (x²)</p>
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<p>The derivative of x/5 is 1/5, and the derivative of x² is 2x. = 1/5 + 2x Therefore, d/dx (x/5 + x²) = 1/5 + 2x.</p>
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<p>The derivative of x/5 is 1/5, and the derivative of x² is 2x. = 1/5 + 2x Therefore, d/dx (x/5 + x²) = 1/5 + 2x.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we differentiate each term of the given function separately and then combine them to obtain the final result.</p>
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<p>In this process, we differentiate each term of the given function separately and then combine them to obtain the final result.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h2>FAQs on the Derivative of x/5</h2>
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<h2>FAQs on the Derivative of x/5</h2>
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<h3>1.Find the derivative of x/5.</h3>
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<h3>1.Find the derivative of x/5.</h3>
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<p>Using the constant rule, the derivative of x/5 is 1/5.</p>
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<p>Using the constant rule, the derivative of x/5 is 1/5.</p>
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<h3>2.Can we use the derivative of x/5 in real life?</h3>
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<h3>2.Can we use the derivative of x/5 in real life?</h3>
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<p>Yes, we can use the derivative of x/5 in real life to calculate the rate of change in cost, speed, or any other linear rate of variation.</p>
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<p>Yes, we can use the derivative of x/5 in real life to calculate the rate of change in cost, speed, or any other linear rate of variation.</p>
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<h3>3.Is the derivative of x/5 affected by changes in x?</h3>
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<h3>3.Is the derivative of x/5 affected by changes in x?</h3>
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<p>No, the derivative of x/5 is constant at 1/5 and does not change with different values of x.</p>
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<p>No, the derivative of x/5 is constant at 1/5 and does not change with different values of x.</p>
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<h3>4.What rule is used to differentiate x/5?</h3>
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<h3>4.What rule is used to differentiate x/5?</h3>
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<p>We use the constant multiple rule to differentiate x/5, resulting in a derivative of 1/5.</p>
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<p>We use the constant multiple rule to differentiate x/5, resulting in a derivative of 1/5.</p>
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<h3>5.Are the derivatives of x/5 and 5x the same?</h3>
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<h3>5.Are the derivatives of x/5 and 5x the same?</h3>
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<p>No, they are different. The derivative of x/5 is 1/5, while the derivative of 5x is 5.</p>
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<p>No, they are different. The derivative of x/5 is 1/5, while the derivative of 5x is 5.</p>
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<h2>Important Glossaries for the Derivative of x/5</h2>
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<h2>Important Glossaries for the Derivative of x/5</h2>
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<ul><li>Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
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<ul><li>Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
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</ul><ul><li>Linear Function: A function of the form f(x) = mx + b, where m and b are constants.</li>
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</ul><ul><li>Linear Function: A function of the form f(x) = mx + b, where m and b are constants.</li>
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</ul><ul><li>Constant Rule: A rule stating that the derivative of a constant multiplied by a function is the constant times the derivative of the function.</li>
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</ul><ul><li>Constant Rule: A rule stating that the derivative of a constant multiplied by a function is the constant times the derivative of the function.</li>
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</ul><ul><li>First Derivative: The initial result of differentiating a function, which gives us the rate of change.</li>
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</ul><ul><li>First Derivative: The initial result of differentiating a function, which gives us the rate of change.</li>
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</ul><ul><li>Constant: A value that does not change, used in the context of the multiple and constant terms in differentiation.</li>
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</ul><ul><li>Constant: A value that does not change, used in the context of the multiple and constant terms in differentiation.</li>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<p>▶</p>
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<p>▶</p>
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<h2>Jaskaran Singh Saluja</h2>
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<h2>Jaskaran Singh Saluja</h2>
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<h3>About the Author</h3>
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<h3>About the Author</h3>
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<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>
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<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>
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<h3>Fun Fact</h3>
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<h3>Fun Fact</h3>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>