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<p>Last updated on<strong>September 22, 2025</strong></p>
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<p>Last updated on<strong>September 22, 2025</strong></p>
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<p>We explore the derivative of sin(xy) using implicit differentiation, which is essential for understanding how the function changes as both x and y vary. This concept is particularly useful in multivariable calculus and real-life applications involving related rates. We will now discuss the derivative of sin(xy) in detail.</p>
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<p>We explore the derivative of sin(xy) using implicit differentiation, which is essential for understanding how the function changes as both x and y vary. This concept is particularly useful in multivariable calculus and real-life applications involving related rates. We will now discuss the derivative of sin(xy) in detail.</p>
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<h2>What is the Derivative of sin(xy)?</h2>
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<h2>What is the Derivative of sin(xy)?</h2>
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<p>To understand the derivative of sin(xy), we need to use implicit differentiation because sin(xy) is a composite<a>function</a>of two<a>variables</a>. It is commonly represented as d/dx (sin(xy)) or (sin(xy))'. The function sin(xy) has a derivative that depends on both x and y, illustrating its differentiability within its domain. The key concepts involve: -</p>
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<p>To understand the derivative of sin(xy), we need to use implicit differentiation because sin(xy) is a composite<a>function</a>of two<a>variables</a>. It is commonly represented as d/dx (sin(xy)) or (sin(xy))'. The function sin(xy) has a derivative that depends on both x and y, illustrating its differentiability within its domain. The key concepts involve: -</p>
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<p><strong>Sine Function:</strong>sin(xy) is a trigonometric function involving the<a>product</a>of x and y. -</p>
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<p><strong>Sine Function:</strong>sin(xy) is a trigonometric function involving the<a>product</a>of x and y. -</p>
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<p><strong>Product Rule:</strong>Necessary for differentiating xy since it's a product of two variables. </p>
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<p><strong>Product Rule:</strong>Necessary for differentiating xy since it's a product of two variables. </p>
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<p><strong>Chain Rule:</strong>Used to differentiate composite functions like sin(xy).</p>
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<p><strong>Chain Rule:</strong>Used to differentiate composite functions like sin(xy).</p>
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<h2>Derivative of sin(xy) Formula</h2>
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<h2>Derivative of sin(xy) Formula</h2>
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<p>The derivative of sin(xy) with respect to x can be expressed using implicit differentiation.</p>
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<p>The derivative of sin(xy) with respect to x can be expressed using implicit differentiation.</p>
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<p>The<a>formula</a>is: d/dx (sin(xy)) = cos(xy) * (y + x (dy/dx))</p>
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<p>The<a>formula</a>is: d/dx (sin(xy)) = cos(xy) * (y + x (dy/dx))</p>
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<p>The formula applies to all x and y where the function is defined and differentiable.</p>
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<p>The formula applies to all x and y where the function is defined and differentiable.</p>
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<h2>Proofs of the Derivative of sin(xy)</h2>
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<h2>Proofs of the Derivative of sin(xy)</h2>
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<p>We can derive the derivative of sin(xy) using implicit differentiation. To show this, we will use trigonometric identities along with differentiation rules. Here are some methods to prove it:</p>
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<p>We can derive the derivative of sin(xy) using implicit differentiation. To show this, we will use trigonometric identities along with differentiation rules. Here are some methods to prove it:</p>
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<h3>Using Implicit Differentiation</h3>
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<h3>Using Implicit Differentiation</h3>
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<p>To find the derivative of sin(xy), we apply implicit differentiation. Assume z = sin(xy).</p>
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<p>To find the derivative of sin(xy), we apply implicit differentiation. Assume z = sin(xy).</p>
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<p>Differentiate both sides with respect to x: d/dx (z) = d/dx (sin(xy)) dz/dx = cos(xy) * (d/dx (xy))</p>
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<p>Differentiate both sides with respect to x: d/dx (z) = d/dx (sin(xy)) dz/dx = cos(xy) * (d/dx (xy))</p>
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<p>Using the product rule on xy: d/dx (xy) = y + x (dy/dx) Thus, dz/dx = cos(xy) * (y + x (dy/dx))</p>
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<p>Using the product rule on xy: d/dx (xy) = y + x (dy/dx) Thus, dz/dx = cos(xy) * (y + x (dy/dx))</p>
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<h3>Using Chain Rule</h3>
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<h3>Using Chain Rule</h3>
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<p>To differentiate sin(xy) using the chain rule, consider u = xy and v = sin(u).</p>
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<p>To differentiate sin(xy) using the chain rule, consider u = xy and v = sin(u).</p>
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<p>Differentiate v with respect to x: dv/dx = cos(u) * du/dx Since u = xy, apply the product rule: du/dx = y + x (dy/dx)</p>
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<p>Differentiate v with respect to x: dv/dx = cos(u) * du/dx Since u = xy, apply the product rule: du/dx = y + x (dy/dx)</p>
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<p>Thus, dv/dx = cos(xy) * (y + x (dy/dx))</p>
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<p>Thus, dv/dx = cos(xy) * (y + x (dy/dx))</p>
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<h3>Using Product Rule</h3>
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<h3>Using Product Rule</h3>
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<p>To differentiate sin(xy) using the product rule, express it as a product: sin(xy) = sin(u), where u = xy.</p>
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<p>To differentiate sin(xy) using the product rule, express it as a product: sin(xy) = sin(u), where u = xy.</p>
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<p>Apply product and chain rules: d/dx (sin(u)) = cos(u) * du/dx du/dx = y + x (dy/dx)</p>
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<p>Apply product and chain rules: d/dx (sin(u)) = cos(u) * du/dx du/dx = y + x (dy/dx)</p>
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<p>Thus, d/dx (sin(xy)) = cos(xy) * (y + x (dy/dx))</p>
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<p>Thus, d/dx (sin(xy)) = cos(xy) * (y + x (dy/dx))</p>
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<h2>Higher-Order Derivatives of sin(xy)</h2>
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<h2>Higher-Order Derivatives of sin(xy)</h2>
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<p>Higher-order derivatives involve differentiating a function<a>multiple</a>times. For functions like sin(xy), understanding higher-order derivatives can be complex but insightful.</p>
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<p>Higher-order derivatives involve differentiating a function<a>multiple</a>times. For functions like sin(xy), understanding higher-order derivatives can be complex but insightful.</p>
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<p>For instance, if sin(xy) describes a wave, the first derivative gives the slope, and the second derivative indicates concavity or convexity.</p>
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<p>For instance, if sin(xy) describes a wave, the first derivative gives the slope, and the second derivative indicates concavity or convexity.</p>
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<p>The nth derivative, denoted as fⁿ(x), provides information about the<a>rate</a>of change at higher levels.</p>
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<p>The nth derivative, denoted as fⁿ(x), provides information about the<a>rate</a>of change at higher levels.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When x or y is such that xy = π/2, the derivative includes cos(π/2), which results in 0, simplifying the<a>expression</a>. When x = 0 or y = 0, the derivative simplifies because sin(0) = 0.</p>
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<p>When x or y is such that xy = π/2, the derivative includes cos(π/2), which results in 0, simplifying the<a>expression</a>. When x = 0 or y = 0, the derivative simplifies because sin(0) = 0.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of sin(xy)</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of sin(xy)</h2>
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<p>Students often make mistakes when differentiating sin(xy). These errors can be rectified by understanding the correct methods. Here are some common mistakes and solutions:</p>
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<p>Students often make mistakes when differentiating sin(xy). These errors can be rectified by understanding the correct methods. Here are some common mistakes and solutions:</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 sin(3xy).</p>
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<p>Calculate the derivative of sin(3xy).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Let z = sin(3xy). Using implicit differentiation, dz/dx = cos(3xy) * d/dx (3xy)</p>
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<p>Let z = sin(3xy). Using implicit differentiation, dz/dx = cos(3xy) * d/dx (3xy)</p>
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<p>Apply the product rule: d/dx (3xy) = 3(y + 3x(dy/dx))</p>
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<p>Apply the product rule: d/dx (3xy) = 3(y + 3x(dy/dx))</p>
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<p>Substitute back: dz/dx = cos(3xy) * 3(y + 3x(dy/dx))</p>
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<p>Substitute back: dz/dx = cos(3xy) * 3(y + 3x(dy/dx))</p>
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<p>Thus, the derivative of sin(3xy) is 3 cos(3xy) * (y + 3x(dy/dx)).</p>
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<p>Thus, the derivative of sin(3xy) is 3 cos(3xy) * (y + 3x(dy/dx)).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the derivative by differentiating sin(3xy) using the chain rule and product rule. This involves differentiating the inner function 3xy and applying the chain rule.</p>
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<p>We find the derivative by differentiating sin(3xy) using the chain rule and product rule. This involves differentiating the inner function 3xy and applying the chain rule.</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 circular track is expanding, and its radius is given by the function r = sin(xy). If x = 2 meters and y = 1 meter, find dr/dx.</p>
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<p>A circular track is expanding, and its radius is given by the function r = sin(xy). If x = 2 meters and y = 1 meter, find dr/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>Given r = sin(xy), differentiate with respect to x: dr/dx = cos(xy) * d/dx (xy)</p>
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<p>Given r = sin(xy), differentiate with respect to x: dr/dx = cos(xy) * d/dx (xy)</p>
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<p>Using the product rule: d/dx (xy) = y + x(dy/dx)</p>
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<p>Using the product rule: d/dx (xy) = y + x(dy/dx)</p>
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<p>Substitute x = 2 and y = 1: dr/dx = cos(2 * 1) * (1 + 2(dy/dx))</p>
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<p>Substitute x = 2 and y = 1: dr/dx = cos(2 * 1) * (1 + 2(dy/dx))</p>
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<p>Since dy/dx is not specified, assume dy/dx = 0 for simplicity: dr/dx = cos(2) * 1</p>
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<p>Since dy/dx is not specified, assume dy/dx = 0 for simplicity: dr/dx = cos(2) * 1</p>
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<p>Thus, dr/dx = cos(2).</p>
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<p>Thus, dr/dx = cos(2).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find dr/dx by differentiating the given function implicitly, substituting given values, and assuming dy/dx = 0 for simplicity.</p>
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<p>We find dr/dx by differentiating the given function implicitly, substituting given values, and assuming dy/dx = 0 for simplicity.</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 z = sin(xy).</p>
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<p>Derive the second derivative of the function z = sin(xy).</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: dz/dx = cos(xy) * (y + x(dy/dx))... (1)</p>
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<p>First, find the first derivative: dz/dx = cos(xy) * (y + x(dy/dx))... (1)</p>
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<p>Now differentiate again: d²z/dx² = d/dx [cos(xy) * (y + x(dy/dx))]</p>
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<p>Now differentiate again: d²z/dx² = d/dx [cos(xy) * (y + x(dy/dx))]</p>
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<p>Apply product and chain rules to (1): = -sin(xy) * (y + x(dy/dx)) * (y + x(dy/dx)) + cos(xy) * (dy/dx + x(d²y/dx²))</p>
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<p>Apply product and chain rules to (1): = -sin(xy) * (y + x(dy/dx)) * (y + x(dy/dx)) + cos(xy) * (dy/dx + x(d²y/dx²))</p>
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<p>Simplify terms: d²z/dx² = -sin(xy) * (y + x(dy/dx))² + cos(xy) * (dy/dx + x(d²y/dx²))</p>
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<p>Simplify terms: d²z/dx² = -sin(xy) * (y + x(dy/dx))² + cos(xy) * (dy/dx + x(d²y/dx²))</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>To derive the second derivative, we apply implicit differentiation to the first derivative, using the product and chain rules to simplify.</p>
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<p>To derive the second derivative, we apply implicit differentiation to the first derivative, using the product and chain rules to simplify.</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 (sin²(xy)) = 2 sin(xy) cos(xy) * (y + x(dy/dx)).</p>
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<p>Prove: d/dx (sin²(xy)) = 2 sin(xy) cos(xy) * (y + x(dy/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>Consider w = sin²(xy). Using the chain rule: dw/dx = 2 sin(xy) * d/dx (sin(xy))</p>
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<p>Consider w = sin²(xy). Using the chain rule: dw/dx = 2 sin(xy) * d/dx (sin(xy))</p>
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<p>Differentiate sin(xy): d/dx (sin(xy)) = cos(xy) * (y + x(dy/dx))</p>
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<p>Differentiate sin(xy): d/dx (sin(xy)) = cos(xy) * (y + x(dy/dx))</p>
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<p>Substitute: dw/dx = 2 sin(xy) cos(xy) * (y + x(dy/dx)) Hence proved.</p>
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<p>Substitute: dw/dx = 2 sin(xy) cos(xy) * (y + x(dy/dx)) Hence proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we apply the chain rule to differentiate sin²(xy) and substitute the derivative of sin(xy) to complete the proof.</p>
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<p>In this process, we apply the chain rule to differentiate sin²(xy) and substitute the derivative of sin(xy) to complete the proof.</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 (sin(xy)/x).</p>
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<p>Solve: d/dx (sin(xy)/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>To differentiate the function, use the quotient rule: d/dx (sin(xy)/x) = (d/dx (sin(xy)) * x - sin(xy) * d/dx(x))/x² Differentiate sin(xy): d/dx (sin(xy)) = cos(xy) * (y + x(dy/dx))</p>
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<p>To differentiate the function, use the quotient rule: d/dx (sin(xy)/x) = (d/dx (sin(xy)) * x - sin(xy) * d/dx(x))/x² Differentiate sin(xy): d/dx (sin(xy)) = cos(xy) * (y + x(dy/dx))</p>
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<p>Substitute: = (cos(xy) * (y + x(dy/dx)) * x - sin(xy))/x² Simplify: = x cos(xy) * (y + x(dy/dx)) - sin(xy)/x²</p>
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<p>Substitute: = (cos(xy) * (y + x(dy/dx)) * x - sin(xy))/x² Simplify: = x cos(xy) * (y + x(dy/dx)) - sin(xy)/x²</p>
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<p>Thus, d/dx (sin(xy)/x) = [x cos(xy) * (y + x(dy/dx)) - sin(xy)]/x²</p>
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<p>Thus, d/dx (sin(xy)/x) = [x cos(xy) * (y + x(dy/dx)) - sin(xy)]/x²</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We differentiate the given function using the quotient rule, simplifying to obtain the final result.</p>
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<p>We differentiate the given function using the quotient rule, simplifying 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 sin(xy)</h2>
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<h2>FAQs on the Derivative of sin(xy)</h2>
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<h3>1.Find the derivative of sin(xy).</h3>
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<h3>1.Find the derivative of sin(xy).</h3>
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<p>Using implicit differentiation on sin(xy), we apply the chain rule: d/dx (sin(xy)) = cos(xy) * (y + x(dy/dx))</p>
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<p>Using implicit differentiation on sin(xy), we apply the chain rule: d/dx (sin(xy)) = cos(xy) * (y + x(dy/dx))</p>
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<h3>2.Can the derivative of sin(xy) be used in real life?</h3>
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<h3>2.Can the derivative of sin(xy) be used in real life?</h3>
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<p>Yes, derivatives<a>of functions</a>like sin(xy) can be applied in real-life scenarios involving rates of change in physics, engineering, and other fields.</p>
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<p>Yes, derivatives<a>of functions</a>like sin(xy) can be applied in real-life scenarios involving rates of change in physics, engineering, and other fields.</p>
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<h3>3.Is it possible to take the derivative of sin(xy) at the point where xy = π/2?</h3>
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<h3>3.Is it possible to take the derivative of sin(xy) at the point where xy = π/2?</h3>
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<p>Yes, at xy = π/2, the cosine term in the derivative becomes 0, greatly simplifying the expression.</p>
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<p>Yes, at xy = π/2, the cosine term in the derivative becomes 0, greatly simplifying the expression.</p>
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<h3>4.What rule is used to differentiate sin(xy)/x?</h3>
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<h3>4.What rule is used to differentiate sin(xy)/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate sin(xy)/x, resulting in: d/dx (sin(xy)/x) = [x cos(xy) * (y + x(dy/dx)) - sin(xy)]/x²</p>
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<p>We use the<a>quotient</a>rule to differentiate sin(xy)/x, resulting in: d/dx (sin(xy)/x) = [x cos(xy) * (y + x(dy/dx)) - sin(xy)]/x²</p>
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<h3>5.Are the derivatives of sin(xy) and sin(yx) the same?</h3>
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<h3>5.Are the derivatives of sin(xy) and sin(yx) the same?</h3>
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<h2>Important Glossaries for the Derivative of sin(xy)</h2>
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<h2>Important Glossaries for the Derivative of sin(xy)</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function describes how the function changes in response to changes in the variables.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function describes how the function changes in response to changes in the variables.</li>
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</ul><ul><li><strong>Implicit Differentiation:</strong>A technique used to find derivatives of functions that are not explicitly solved for one variable in terms of another.</li>
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</ul><ul><li><strong>Implicit Differentiation:</strong>A technique used to find derivatives of functions that are not explicitly solved for one variable in terms of another.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate products of two functions.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate products of two functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used for differentiating composite functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used for differentiating composite functions.</li>
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</ul><ul><li><strong>Trigonometric Functions:</strong>Functions like sine and cosine that are fundamental in calculus and analysis.</li>
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</ul><ul><li><strong>Trigonometric Functions:</strong>Functions like sine and cosine that are fundamental in calculus and analysis.</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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<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>