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2026-01-01
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<p>Last updated on<strong>September 27, 2025</strong></p>
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<p>Last updated on<strong>September 27, 2025</strong></p>
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<p>We use the derivative of sin²(2x) to understand how this composite trigonometric function changes with respect to x. Derivatives help us in various real-life applications, including physics and engineering. We will investigate the derivative of sin²(2x) in detail.</p>
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<p>We use the derivative of sin²(2x) to understand how this composite trigonometric function changes with respect to x. Derivatives help us in various real-life applications, including physics and engineering. We will investigate the derivative of sin²(2x) in detail.</p>
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<h2>What is the Derivative of sin²(2x)?</h2>
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<h2>What is the Derivative of sin²(2x)?</h2>
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<p>We now explore the derivative<a>of</a>sin²(2x). It is commonly represented as d/dx (sin²(2x)) or (sin²(2x))'. The derivative of sin²(2x) involves using the chain rule and the<a>power</a>rule. The<a>function</a>sin²(2x) is differentiable within its domain.</p>
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<p>We now explore the derivative<a>of</a>sin²(2x). It is commonly represented as d/dx (sin²(2x)) or (sin²(2x))'. The derivative of sin²(2x) involves using the chain rule and the<a>power</a>rule. The<a>function</a>sin²(2x) is differentiable within its domain.</p>
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<p>Key concepts include:</p>
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<p>Key concepts include:</p>
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<p><strong>Sine Function:</strong>sin(2x) is the inner function.</p>
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<p><strong>Sine Function:</strong>sin(2x) is the inner function.</p>
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<p><strong>Chain Rule:</strong>Used for differentiating composite functions like sin²(2x).</p>
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<p><strong>Chain Rule:</strong>Used for differentiating composite functions like sin²(2x).</p>
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<p><strong>Power Rule:</strong>Used for the<a>exponent</a>part of the function.</p>
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<p><strong>Power Rule:</strong>Used for the<a>exponent</a>part of the function.</p>
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<h2>Derivative of sin²(2x) Formula</h2>
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<h2>Derivative of sin²(2x) Formula</h2>
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<p>The derivative of sin²(2x) can be denoted as d/dx (sin²(2x)). To find it, we apply the chain rule and the power rule: d/dx (sin²(2x)) = 2 * sin(2x) * d/dx(sin(2x))</p>
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<p>The derivative of sin²(2x) can be denoted as d/dx (sin²(2x)). To find it, we apply the chain rule and the power rule: d/dx (sin²(2x)) = 2 * sin(2x) * d/dx(sin(2x))</p>
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<p>Next, apply the chain rule on sin(2x): d/dx(sin(2x)) = 2 * cos(2x) Thus, the derivative is: d/dx (sin²(2x)) = 4 * sin(2x) * cos(2x)</p>
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<p>Next, apply the chain rule on sin(2x): d/dx(sin(2x)) = 2 * cos(2x) Thus, the derivative is: d/dx (sin²(2x)) = 4 * sin(2x) * cos(2x)</p>
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<h2>Proofs of the Derivative of sin²(2x)</h2>
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<h2>Proofs of the Derivative of sin²(2x)</h2>
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<p>We can derive the derivative of sin²(2x) using proofs. This involves applying differentiation rules and trigonometric identities.</p>
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<p>We can derive the derivative of sin²(2x) using proofs. This involves applying differentiation rules and trigonometric identities.</p>
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<p>Here are a few methods: </p>
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<p>Here are a few methods: </p>
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<ul><li>By Chain Rule </li>
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<ul><li>By Chain Rule </li>
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<li>By Trigonometric Identities</li>
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<li>By Trigonometric Identities</li>
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</ul><p>We will demonstrate that the differentiation of sin²(2x) results in 4 * sin(2x) * cos(2x) using these methods:</p>
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</ul><p>We will demonstrate that the differentiation of sin²(2x) results in 4 * sin(2x) * cos(2x) using these methods:</p>
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<h2><strong>By Chain Rule</strong></h2>
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<h2><strong>By Chain Rule</strong></h2>
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<p>To prove using the chain rule, consider u = sin(2x), then y = u². Differentiate y = u² with respect to u: dy/du = 2u Differentiate u = sin(2x) with respect to x: du/dx = 2cos(2x) Using the chain rule: dy/dx = dy/du * du/dx = 2u * 2cos(2x) Substitute u = sin(2x): dy/dx = 2 * sin(2x) * 2cos(2x) = 4 * sin(2x) * cos(2x)</p>
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<p>To prove using the chain rule, consider u = sin(2x), then y = u². Differentiate y = u² with respect to u: dy/du = 2u Differentiate u = sin(2x) with respect to x: du/dx = 2cos(2x) Using the chain rule: dy/dx = dy/du * du/dx = 2u * 2cos(2x) Substitute u = sin(2x): dy/dx = 2 * sin(2x) * 2cos(2x) = 4 * sin(2x) * cos(2x)</p>
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<h2><strong>By Trigonometric Identities</strong></h2>
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<h2><strong>By Trigonometric Identities</strong></h2>
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<p>We know sin²(2x) = (sin(2x))². Using the identity sin(2x) = 2sin(x)cos(x), we differentiate: d/dx (sin²(2x)) = d/dx (2sin(x)cos(x))² Using the<a>product</a>rule and chain rule, we find: d/dx (sin²(2x)) = 4sin(2x)cos(2x)</p>
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<p>We know sin²(2x) = (sin(2x))². Using the identity sin(2x) = 2sin(x)cos(x), we differentiate: d/dx (sin²(2x)) = d/dx (2sin(x)cos(x))² Using the<a>product</a>rule and chain rule, we find: d/dx (sin²(2x)) = 4sin(2x)cos(2x)</p>
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<h3>Explore Our Programs</h3>
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<h2>Higher-Order Derivatives of sin²(2x)</h2>
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<h2>Higher-Order Derivatives of sin²(2x)</h2>
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<p>When a function is differentiated<a>multiple</a>times, the resulting derivatives are referred to as higher-order derivatives.</p>
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<p>When a function is differentiated<a>multiple</a>times, the resulting derivatives are referred to as higher-order derivatives.</p>
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<p>For sin²(2x), the first derivative is f′(x), indicating how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x).</p>
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<p>For sin²(2x), the first derivative is f′(x), indicating how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x).</p>
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<p>The third derivative, f′′′(x), follows from the second derivative, and this pattern continues. The nth derivative of sin²(2x) is generally denoted as fⁿ(x), representing the change in the<a>rate</a>of change.</p>
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<p>The third derivative, f′′′(x), follows from the second derivative, and this pattern continues. The nth derivative of sin²(2x) is generally denoted as fⁿ(x), representing the change in the<a>rate</a>of change.</p>
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<h2>Special Cases</h2>
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<h2>Special Cases</h2>
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<p>When x is 0, the derivative of sin²(2x) = 4 * sin(0) * cos(0) = 0.</p>
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<p>When x is 0, the derivative of sin²(2x) = 4 * sin(0) * cos(0) = 0.</p>
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<p>At points where sin(2x) = 0, such as x = π, 2π, ..., the derivative is also 0.</p>
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<p>At points where sin(2x) = 0, such as x = π, 2π, ..., the derivative is also 0.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of sin²(2x)</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of sin²(2x)</h2>
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<p>Students frequently make mistakes when differentiating sin²(2x). 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 sin²(2x). 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 sin²(2x) · cos²(2x)</p>
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<p>Calculate the derivative of sin²(2x) · cos²(2x)</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) = sin²(2x) · cos²(2x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin²(2x) and v = cos²(2x). Let’s differentiate each term, u′ = d/dx (sin²(2x)) = 4sin(2x)cos(2x) v′ = d/dx (cos²(2x)) = -4cos(2x)sin(2x) Substituting into the given equation, f'(x) = (4sin(2x)cos(2x)) · (cos²(2x)) + (sin²(2x)) · (-4cos(2x)sin(2x)) Simplifying terms, we get, f'(x) = 4sin(2x)cos³(2x) - 4sin³(2x)cos(2x) Thus, the derivative of the specified function is 4sin(2x)cos(2x)(cos²(2x) - sin²(2x)).</p>
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<p>Here, we have f(x) = sin²(2x) · cos²(2x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin²(2x) and v = cos²(2x). Let’s differentiate each term, u′ = d/dx (sin²(2x)) = 4sin(2x)cos(2x) v′ = d/dx (cos²(2x)) = -4cos(2x)sin(2x) Substituting into the given equation, f'(x) = (4sin(2x)cos(2x)) · (cos²(2x)) + (sin²(2x)) · (-4cos(2x)sin(2x)) Simplifying terms, we get, f'(x) = 4sin(2x)cos³(2x) - 4sin³(2x)cos(2x) Thus, the derivative of the specified function is 4sin(2x)cos(2x)(cos²(2x) - sin²(2x)).</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 dividing it into two parts.</p>
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<p>We find the derivative of the given function by dividing it into two parts.</p>
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<p>The first step is finding the derivative of each part and then combining them using the product rule to get the final result.</p>
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<p>The first step is finding the derivative of each part and then combining them using the product rule 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 pendulum swings with its angular position modeled by θ = sin²(2x), where x is time in seconds. Find the rate of change of the angle at x = π/6 seconds.</p>
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<p>A pendulum swings with its angular position modeled by θ = sin²(2x), where x is time in seconds. Find the rate of change of the angle at x = π/6 seconds.</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 θ = sin²(2x) (angular position of the pendulum)...(1) Now, we will differentiate equation (1). Take the derivative of sin²(2x): dθ/dx = 4sin(2x)cos(2x) Given x = π/6, substitute this into the derivative: dθ/dx = 4sin(π/3)cos(π/3) Using values sin(π/3) = √3/2 and cos(π/3) = 1/2: dθ/dx = 4(√3/2)(1/2) = √3 Hence, the rate of change of the angle at x = π/6 seconds is √3.</p>
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<p>We have θ = sin²(2x) (angular position of the pendulum)...(1) Now, we will differentiate equation (1). Take the derivative of sin²(2x): dθ/dx = 4sin(2x)cos(2x) Given x = π/6, substitute this into the derivative: dθ/dx = 4sin(π/3)cos(π/3) Using values sin(π/3) = √3/2 and cos(π/3) = 1/2: dθ/dx = 4(√3/2)(1/2) = √3 Hence, the rate of change of the angle at x = π/6 seconds is √3.</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 the pendulum’s angle at x = π/6 seconds by substituting the value into the derivative.</p>
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<p>We find the rate of change of the pendulum’s angle at x = π/6 seconds by substituting the value into the derivative.</p>
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<p>This gives the instantaneous rate of change of the angle with respect to time.</p>
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<p>This gives the instantaneous rate of change of the angle with respect to time.</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 = sin²(2x).</p>
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<p>Derive the second derivative of the function y = sin²(2x).</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 = 4sin(2x)cos(2x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [4sin(2x)cos(2x)] Using the product rule, d²y/dx² = 4[d/dx(sin(2x)cos(2x))] = 4[2cos²(2x) - 2sin²(2x)] = 8[cos²(2x) - sin²(2x)] Therefore, the second derivative of the function y = sin²(2x) is 8[cos²(2x) - sin²(2x)].</p>
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<p>The first step is to find the first derivative, dy/dx = 4sin(2x)cos(2x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [4sin(2x)cos(2x)] Using the product rule, d²y/dx² = 4[d/dx(sin(2x)cos(2x))] = 4[2cos²(2x) - 2sin²(2x)] = 8[cos²(2x) - sin²(2x)] Therefore, the second derivative of the function y = sin²(2x) is 8[cos²(2x) - sin²(2x)].</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>Using the product rule, we differentiate sin(2x)cos(2x).</p>
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<p>Using the product rule, we differentiate sin(2x)cos(2x).</p>
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<p>We simplify the terms to find the final answer for the second derivative.</p>
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<p>We simplify the terms to find the final answer for the second 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 (sin²(2x)) = 4sin(2x)cos(2x).</p>
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<p>Prove: d/dx (sin²(2x)) = 4sin(2x)cos(2x).</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’s start using the chain rule: Consider y = sin²(2x). Let u = sin(2x), then y = u². To differentiate, use the chain rule: dy/dx = 2u * d/dx(u) Since d/dx(sin(2x)) = 2cos(2x), dy/dx = 2sin(2x) * 2cos(2x) = 4sin(2x)cos(2x) Hence proved.</p>
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<p>Let’s start using the chain rule: Consider y = sin²(2x). Let u = sin(2x), then y = u². To differentiate, use the chain rule: dy/dx = 2u * d/dx(u) Since d/dx(sin(2x)) = 2cos(2x), dy/dx = 2sin(2x) * 2cos(2x) = 4sin(2x)cos(2x) 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 chain rule to differentiate the equation.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation.</p>
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<p>We replace sin(2x) with u, differentiate, and substitute back to prove the result.</p>
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<p>We replace sin(2x) with u, differentiate, and substitute back to prove 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 (sin²(2x)/ x)</p>
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<p>Solve: d/dx (sin²(2x)/ 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, we use the quotient rule: d/dx (sin²(2x)/ x) = (d/dx (sin²(2x)) * x - sin²(2x) * d/dx(x)) / x² Substitute d/dx(sin²(2x)) = 4sin(2x)cos(2x) and d/dx(x) = 1: = (4sin(2x)cos(2x) * x - sin²(2x)) / x² = (4xsin(2x)cos(2x) - sin²(2x)) / x² Therefore, d/dx (sin²(2x)/ x) = (4xsin(2x)cos(2x) - sin²(2x)) / x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (sin²(2x)/ x) = (d/dx (sin²(2x)) * x - sin²(2x) * d/dx(x)) / x² Substitute d/dx(sin²(2x)) = 4sin(2x)cos(2x) and d/dx(x) = 1: = (4sin(2x)cos(2x) * x - sin²(2x)) / x² = (4xsin(2x)cos(2x) - sin²(2x)) / x² Therefore, d/dx (sin²(2x)/ x) = (4xsin(2x)cos(2x) - sin²(2x)) / x²</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 the given function using the quotient rule.</p>
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<p>In this process, we differentiate the given function using the quotient rule.</p>
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<p>As a final step, we simplify the equation to obtain the final result.</p>
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<p>As a final step, we simplify the equation 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 sin²(2x)</h2>
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<h2>FAQs on the Derivative of sin²(2x)</h2>
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<h3>1.Find the derivative of sin²(2x).</h3>
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<h3>1.Find the derivative of sin²(2x).</h3>
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<p>Using the chain rule and the power rule: d/dx (sin²(2x)) = 4sin(2x)cos(2x)</p>
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<p>Using the chain rule and the power rule: d/dx (sin²(2x)) = 4sin(2x)cos(2x)</p>
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<h3>2.Can we use the derivative of sin²(2x) in real life?</h3>
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<h3>2.Can we use the derivative of sin²(2x) in real life?</h3>
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<p>Yes, the derivative of sin²(2x) can be used in real-life applications, such as analyzing waveforms in physics or engineering contexts.</p>
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<p>Yes, the derivative of sin²(2x) can be used in real-life applications, such as analyzing waveforms in physics or engineering contexts.</p>
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<h3>3.Is it possible to take the derivative of sin²(2x) at x = π?</h3>
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<h3>3.Is it possible to take the derivative of sin²(2x) at x = π?</h3>
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<p>Yes, it is possible to take the derivative at x = π since the function sin²(2x) is differentiable at this point and its derivative is 0.</p>
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<p>Yes, it is possible to take the derivative at x = π since the function sin²(2x) is differentiable at this point and its derivative is 0.</p>
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<h3>4.What rule is used to differentiate sin²(2x)/x?</h3>
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<h3>4.What rule is used to differentiate sin²(2x)/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate sin²(2x)/x. The derivative is (4xsin(2x)cos(2x) - sin²(2x)) / x².</p>
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<p>We use the<a>quotient</a>rule to differentiate sin²(2x)/x. The derivative is (4xsin(2x)cos(2x) - sin²(2x)) / x².</p>
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<h3>5.Are the derivatives of sin²(2x) and sin²(x) the same?</h3>
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<h3>5.Are the derivatives of sin²(2x) and sin²(x) the same?</h3>
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<p>No, they are different. The derivative of sin²(2x) is 4sin(2x)cos(2x), while the derivative of sin²(x) is 2sin(x)cos(x).</p>
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<p>No, they are different. The derivative of sin²(2x) is 4sin(2x)cos(2x), while the derivative of sin²(x) is 2sin(x)cos(x).</p>
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<h2>Important Glossaries for the Derivative of sin²(2x)</h2>
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<h2>Important Glossaries for the Derivative of sin²(2x)</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the function changes with a small change in x.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the function changes with a small change in x.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate composite functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate composite functions.</li>
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</ul><ul><li><strong>Power Rule:</strong>A rule used to differentiate functions of the form u^n.</li>
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</ul><ul><li><strong>Power Rule:</strong>A rule used to differentiate functions of the form u^n.</li>
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</ul><ul><li><strong>Sine Function:</strong>A trigonometric function representing the y-coordinate of a point on the unit circle.</li>
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</ul><ul><li><strong>Sine Function:</strong>A trigonometric function representing the y-coordinate of a point on the unit circle.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate the product of two functions.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate the product of two functions.</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>