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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 3xy to understand how the function changes with respect to changes in x. Derivatives are crucial tools for calculating rates of change in various real-life scenarios. In this section, we will delve into the derivative of 3xy and its implications.</p>
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<p>We use the derivative of 3xy to understand how the function changes with respect to changes in x. Derivatives are crucial tools for calculating rates of change in various real-life scenarios. In this section, we will delve into the derivative of 3xy and its implications.</p>
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<h2>What is the Derivative of 3xy?</h2>
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<h2>What is the Derivative of 3xy?</h2>
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<p>The derivative<a>of</a>the<a>function</a>3xy with respect to x is found using the<a>product</a>rule of differentiation. The function 3xy represents a product of 3x and y, and its derivative can be denoted as d/dx (3xy). The key concepts to understand here include: - Product Rule: Used to differentiate products of two functions. - Constant Multiplication Rule: The derivative of a<a>constant</a>times a function is the constant times the derivative of the function.</p>
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<p>The derivative<a>of</a>the<a>function</a>3xy with respect to x is found using the<a>product</a>rule of differentiation. The function 3xy represents a product of 3x and y, and its derivative can be denoted as d/dx (3xy). The key concepts to understand here include: - Product Rule: Used to differentiate products of two functions. - Constant Multiplication Rule: The derivative of a<a>constant</a>times a function is the constant times the derivative of the function.</p>
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<h2>Derivative of 3xy Formula</h2>
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<h2>Derivative of 3xy Formula</h2>
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<p>To differentiate 3xy with respect to x, we apply the product rule. If y is a function of x, the derivative of 3xy is: d/dx (3xy) = 3(dy/dx)x + 3y This<a>formula</a>applies to all x where y is differentiable with respect to x.</p>
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<p>To differentiate 3xy with respect to x, we apply the product rule. If y is a function of x, the derivative of 3xy is: d/dx (3xy) = 3(dy/dx)x + 3y This<a>formula</a>applies to all x where y is differentiable with respect to x.</p>
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<h2>Proofs of the Derivative of 3xy</h2>
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<h2>Proofs of the Derivative of 3xy</h2>
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<p>We can derive the derivative of 3xy using the product rule. To demonstrate this, the following methods are used: - By Applying the Product Rule - Using Constant Multiplication Rule By Applying the Product Rule: The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. Consider f(x) = 3x and g(x) = y. Thus, d/dx (3xy) = 3y + 3x(dy/dx) Using Constant Multiplication Rule: The constant<a>multiplication</a>rule states that a constant<a>factor</a>can be taken out of the differentiation operation. Thus, d/dx (3xy) = 3(d/dx (xy)) Applying the product rule on xy, we get: d/dx (xy) = y + x(dy/dx) Therefore, d/dx (3xy) = 3(y + x(dy/dx))</p>
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<p>We can derive the derivative of 3xy using the product rule. To demonstrate this, the following methods are used: - By Applying the Product Rule - Using Constant Multiplication Rule By Applying the Product Rule: The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. Consider f(x) = 3x and g(x) = y. Thus, d/dx (3xy) = 3y + 3x(dy/dx) Using Constant Multiplication Rule: The constant<a>multiplication</a>rule states that a constant<a>factor</a>can be taken out of the differentiation operation. Thus, d/dx (3xy) = 3(d/dx (xy)) Applying the product rule on xy, we get: d/dx (xy) = y + x(dy/dx) Therefore, d/dx (3xy) = 3(y + x(dy/dx))</p>
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<h2>Higher-Order Derivatives of 3xy</h2>
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<h2>Higher-Order Derivatives of 3xy</h2>
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<p>Higher-order derivatives involve differentiating a function<a>multiple</a>times. In the context of 3xy, the first derivative provides the<a>rate</a>of change, while the second derivative gives the rate of change of the rate of change, and so on. For the first derivative, we write f′(x), which indicates how the function changes. The second derivative, f″(x), is derived from the first derivative and gives insight into the curvature or concavity of the function. Similarly, the third derivative, f‴(x), is the derivative of the second derivative, and this pattern continues.</p>
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<p>Higher-order derivatives involve differentiating a function<a>multiple</a>times. In the context of 3xy, the first derivative provides the<a>rate</a>of change, while the second derivative gives the rate of change of the rate of change, and so on. For the first derivative, we write f′(x), which indicates how the function changes. The second derivative, f″(x), is derived from the first derivative and gives insight into the curvature or concavity of the function. Similarly, the third derivative, f‴(x), is the derivative of the second derivative, and this pattern continues.</p>
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<h2>Special Cases</h2>
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<h2>Special Cases</h2>
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<p>In some cases, specific values of x may lead to undefined derivatives or special conditions. - If y is constant, the derivative simplifies significantly. - If x or y is zero, the derivative may simplify to zero or another specific value.</p>
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<p>In some cases, specific values of x may lead to undefined derivatives or special conditions. - If y is constant, the derivative simplifies significantly. - If x or y is zero, the derivative may simplify to zero or another specific value.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 3xy</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 3xy</h2>
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<p>Students often make errors when differentiating 3xy. These mistakes can be avoided by understanding the correct procedures. Here are some common mistakes and how to solve them:</p>
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<p>Students often make errors when differentiating 3xy. These mistakes can be avoided by understanding the correct procedures. Here are some common mistakes and how 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 3xy if y = x².</p>
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<p>Calculate the derivative of 3xy if y = 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>Given y = x², the expression becomes f(x) = 3x(x²). Using the product rule, f'(x) = 3(d/dx(x²))x + 3x² = 3(2x)x + 3x² = 6x² + 3x² = 9x²</p>
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<p>Given y = x², the expression becomes f(x) = 3x(x²). Using the product rule, f'(x) = 3(d/dx(x²))x + 3x² = 3(2x)x + 3x² = 6x² + 3x² = 9x²</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We used the product rule to differentiate 3xy with y = x². By substituting y and applying the rule, we found the derivative and simplified the expression to get the final result.</p>
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<p>We used the product rule to differentiate 3xy with y = x². By substituting y and applying the rule, we found the derivative and simplified the expression to get the final result.</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 car travels on a road where its position is given by the function s = 3xy, where y is the velocity at time t. If y = 5t, find the derivative at t = 2.</p>
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<p>A car travels on a road where its position is given by the function s = 3xy, where y is the velocity at time t. If y = 5t, find the derivative at t = 2.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Given y = 5t, s = 3x(5t). Using the product rule, ds/dt = 3(d/dt(5t))x + 3(5t)dx/dt = 3(5)x + 3(5t)dx/dt At t = 2, dx/dt = 0 (assuming x is constant for simplicity), ds/dt = 15x + 15(2) = 15x + 30</p>
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<p>Given y = 5t, s = 3x(5t). Using the product rule, ds/dt = 3(d/dt(5t))x + 3(5t)dx/dt = 3(5)x + 3(5t)dx/dt At t = 2, dx/dt = 0 (assuming x is constant for simplicity), ds/dt = 15x + 15(2) = 15x + 30</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We used the product rule to find the derivative of the position function in terms of time. Substituting t = 2 and simplifying gave us the rate of change of the position at that moment.</p>
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<p>We used the product rule to find the derivative of the position function in terms of time. Substituting t = 2 and simplifying gave us the rate of change of the position at that moment.</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>Determine the second derivative of f(x) = 3xy, where y = ln(x).</p>
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<p>Determine the second derivative of f(x) = 3xy, where y = ln(x).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>First, find the first derivative: f'(x) = 3(d/dx(ln(x)))x + 3ln(x) = 3(1/x)x + 3ln(x) = 3 + 3ln(x) Now find the second derivative: f″(x) = d/dx(3 + 3ln(x)) = 0 + 3(1/x) = 3/x</p>
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<p>First, find the first derivative: f'(x) = 3(d/dx(ln(x)))x + 3ln(x) = 3(1/x)x + 3ln(x) = 3 + 3ln(x) Now find the second derivative: f″(x) = d/dx(3 + 3ln(x)) = 0 + 3(1/x) = 3/x</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We first determined the first derivative using the product rule and the derivative of ln(x). Then, we found the second derivative by differentiating the simplified expression of the first derivative.</p>
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<p>We first determined the first derivative using the product rule and the derivative of ln(x). Then, we found the second derivative by differentiating the simplified expression of the first derivative.</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 (3x²y) = 6xy + 3x²dy/dx</p>
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<p>Prove: d/dx (3x²y) = 6xy + 3x²dy/dx</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Using the product rule, let u = 3x² and v = y. Then, d/dx (3x²y) = u'v + uv' = (6x)y + 3x²(dy/dx) = 6xy + 3x²dy/dx</p>
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<p>Using the product rule, let u = 3x² and v = y. Then, d/dx (3x²y) = u'v + uv' = (6x)y + 3x²(dy/dx) = 6xy + 3x²dy/dx</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We applied the product rule to the expression 3x²y, differentiating each part separately and combining the results to prove the derivative formula.</p>
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<p>We applied the product rule to the expression 3x²y, differentiating each part separately and combining the results to prove the derivative formula.</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 (3xy²) where y = x³.</p>
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<p>Solve: d/dx (3xy²) where y = x³.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Given y = x³, we need to differentiate 3x(x³)². First, simplify to 3x(x⁶) = 3x⁷. Now differentiate: d/dx (3x⁷) = 21x⁶</p>
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<p>Given y = x³, we need to differentiate 3x(x³)². First, simplify to 3x(x⁶) = 3x⁷. Now differentiate: d/dx (3x⁷) = 21x⁶</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We substituted y = x³ into 3xy², simplified the expression, and then differentiated using the power rule to find the solution.</p>
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<p>We substituted y = x³ into 3xy², simplified the expression, and then differentiated using the power rule to find the solution.</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 3xy</h2>
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<h2>FAQs on the Derivative of 3xy</h2>
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<h3>1.Find the derivative of 3xy.</h3>
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<h3>1.Find the derivative of 3xy.</h3>
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<p>Using the product rule, d/dx (3xy) = 3y + 3x(dy/dx)</p>
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<p>Using the product rule, d/dx (3xy) = 3y + 3x(dy/dx)</p>
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<h3>2.Can the derivative of 3xy be used in practical applications?</h3>
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<h3>2.Can the derivative of 3xy be used in practical applications?</h3>
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<p>Yes, derivatives like 3xy are used in physics, engineering, and economics to model and predict changes in systems.</p>
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<p>Yes, derivatives like 3xy are used in physics, engineering, and economics to model and predict changes in systems.</p>
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<h3>3.What if y is constant in 3xy?</h3>
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<h3>3.What if y is constant in 3xy?</h3>
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<p>If y is constant, the derivative simplifies to 3y, as dy/dx would be zero.</p>
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<p>If y is constant, the derivative simplifies to 3y, as dy/dx would be zero.</p>
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<h3>4.How does the product rule apply to 3xy?</h3>
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<h3>4.How does the product rule apply to 3xy?</h3>
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<p>The product rule states that the derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.</p>
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<p>The product rule states that the derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.</p>
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<h3>5.Are the derivatives of 3xy and 3x/y the same?</h3>
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<h3>5.Are the derivatives of 3xy and 3x/y the same?</h3>
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<p>No, they are different. 3xy and 3x/y are different functions, and their derivatives are calculated using different rules (product vs.<a>quotient</a>).</p>
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<p>No, they are different. 3xy and 3x/y are different functions, and their derivatives are calculated using different rules (product vs.<a>quotient</a>).</p>
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<h2>Important Glossaries for the Derivative of 3xy</h2>
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<h2>Important Glossaries for the Derivative of 3xy</h2>
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<p>Derivative: A measure of how a function changes as its input changes. Product Rule: A rule used to differentiate products of two functions. Constant Multiplication Rule: The derivative of a constant times a function is the constant times the derivative of the function. Higher-Order Derivatives: Derivatives of derivatives, giving additional insights into the behavior of functions. Chain Rule: A rule for differentiating compositions of functions.</p>
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<p>Derivative: A measure of how a function changes as its input changes. Product Rule: A rule used to differentiate products of two functions. Constant Multiplication Rule: The derivative of a constant times a function is the constant times the derivative of the function. Higher-Order Derivatives: Derivatives of derivatives, giving additional insights into the behavior of functions. Chain Rule: A rule for differentiating compositions of functions.</p>
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<p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</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>