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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 the function 2sinxcosx, which simplifies to sin(2x), to understand how this function changes in response to a slight change in x. Derivatives help us in various real-life applications such as physics and engineering. We will now discuss the derivative of 2sinxcosx in detail.</p>
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<p>We use the derivative of the function 2sinxcosx, which simplifies to sin(2x), to understand how this function changes in response to a slight change in x. Derivatives help us in various real-life applications such as physics and engineering. We will now discuss the derivative of 2sinxcosx in detail.</p>
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<h2>What is the Derivative of 2sinxcosx?</h2>
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<h2>What is the Derivative of 2sinxcosx?</h2>
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<p>We now understand the derivative<a>of</a>2sinxcosx. This<a>function</a>simplifies to sin(2x), which is differentiable within its domain. The derivative of sin(2x) is commonly represented as d/dx (sin(2x)) or (sin(2x))', and its value is 2cos(2x). The key concepts are mentioned below: - Double Angle Formula: 2sinxcosx = sin(2x). - Derivative of Sine: The derivative of sin(x) is cos(x). - Chain Rule: Rule for differentiating composite functions like sin(2x).</p>
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<p>We now understand the derivative<a>of</a>2sinxcosx. This<a>function</a>simplifies to sin(2x), which is differentiable within its domain. The derivative of sin(2x) is commonly represented as d/dx (sin(2x)) or (sin(2x))', and its value is 2cos(2x). The key concepts are mentioned below: - Double Angle Formula: 2sinxcosx = sin(2x). - Derivative of Sine: The derivative of sin(x) is cos(x). - Chain Rule: Rule for differentiating composite functions like sin(2x).</p>
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<h2>Derivative of 2sinxcosx Formula</h2>
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<h2>Derivative of 2sinxcosx Formula</h2>
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<p>The derivative of 2sinxcosx can be denoted as d/dx (2sinxcosx) or (2sinxcosx)'. The<a>formula</a>we use to differentiate 2sinxcosx, which simplifies to sin(2x), is: \[d/dx (sin(2x)) = 2cos(2x)\] The formula applies to all x within the domain of the cosine function.</p>
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<p>The derivative of 2sinxcosx can be denoted as d/dx (2sinxcosx) or (2sinxcosx)'. The<a>formula</a>we use to differentiate 2sinxcosx, which simplifies to sin(2x), is: \[d/dx (sin(2x)) = 2cos(2x)\] The formula applies to all x within the domain of the cosine function.</p>
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<h2>Proofs of the Derivative of 2sinxcosx</h2>
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<h2>Proofs of the Derivative of 2sinxcosx</h2>
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<p>We can derive the derivative of 2sinxcosx using proofs. To show this, we will use trigonometric identities along with differentiation rules. Several methods can be used to prove this, such as: - Using the Double Angle Identity - Applying the Chain Rule - Using the Product Rule We will demonstrate that the differentiation of 2sinxcosx results in 2cos(2x) using these methods: Using the Double Angle Identity The function 2sinxcosx can be rewritten using the double angle identity: \[2sinxcosx = sin(2x)\] The derivative of sin(2x) is found using the chain rule: \[d/dx (sin(2x)) = 2cos(2x)\] Applying the Chain Rule To differentiate sin(2x) using the chain rule, let: \[u = 2x\] \[d/dx (sin(u)) = cos(u) \cdot du/dx = cos(2x) \cdot 2 = 2cos(2x)\] Using the Product Rule For the<a>product</a>2sinxcosx, let: \[u = 2sinx \quad \text{and} \quad v = cosx\] \[d/dx (u \cdot v) = u' \cdot v + u \cdot v'\] \[u' = 2cosx, \quad v' = -sinx\] \[d/dx (2sinxcosx) = 2cosx \cdot cosx + 2sinx \cdot (-sinx)\] \[= 2cos^2x - 2sin^2x = 2(cos^2x - sin^2x) = 2cos(2x)\]</p>
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<p>We can derive the derivative of 2sinxcosx using proofs. To show this, we will use trigonometric identities along with differentiation rules. Several methods can be used to prove this, such as: - Using the Double Angle Identity - Applying the Chain Rule - Using the Product Rule We will demonstrate that the differentiation of 2sinxcosx results in 2cos(2x) using these methods: Using the Double Angle Identity The function 2sinxcosx can be rewritten using the double angle identity: \[2sinxcosx = sin(2x)\] The derivative of sin(2x) is found using the chain rule: \[d/dx (sin(2x)) = 2cos(2x)\] Applying the Chain Rule To differentiate sin(2x) using the chain rule, let: \[u = 2x\] \[d/dx (sin(u)) = cos(u) \cdot du/dx = cos(2x) \cdot 2 = 2cos(2x)\] Using the Product Rule For the<a>product</a>2sinxcosx, let: \[u = 2sinx \quad \text{and} \quad v = cosx\] \[d/dx (u \cdot v) = u' \cdot v + u \cdot v'\] \[u' = 2cosx, \quad v' = -sinx\] \[d/dx (2sinxcosx) = 2cosx \cdot cosx + 2sinx \cdot (-sinx)\] \[= 2cos^2x - 2sin^2x = 2(cos^2x - sin^2x) = 2cos(2x)\]</p>
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<h2>Higher-Order Derivatives of 2sinxcosx</h2>
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<h2>Higher-Order Derivatives of 2sinxcosx</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. Higher-order derivatives can be complex. For instance, consider a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to analyze functions like 2sinxcosx. For the first derivative of a function, we write f′(x), indicating the function's rate of change or slope at a point. The second derivative is derived from the first derivative, denoted as f′′(x). Similarly, the third derivative, f′′′(x), is derived from the second derivative, and this pattern continues. For the nth Derivative of 2sinxcosx, we use fⁿ(x) to denote the nth derivative of a function f(x), describing the change in the rate of change.</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. Higher-order derivatives can be complex. For instance, consider a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to analyze functions like 2sinxcosx. For the first derivative of a function, we write f′(x), indicating the function's rate of change or slope at a point. The second derivative is derived from the first derivative, denoted as f′′(x). Similarly, the third derivative, f′′′(x), is derived from the second derivative, and this pattern continues. For the nth Derivative of 2sinxcosx, we use fⁿ(x) to denote the nth derivative of a function f(x), describing the change 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>At x = π/2, the derivative of 2sinxcosx is 2cos(π) = -2. At x = 0, the derivative of 2sinxcosx = 2cos(0) = 2.</p>
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<p>At x = π/2, the derivative of 2sinxcosx is 2cos(π) = -2. At x = 0, the derivative of 2sinxcosx = 2cos(0) = 2.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 2sinxcosx</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 2sinxcosx</h2>
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<p>Students frequently make mistakes when differentiating 2sinxcosx. 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 2sinxcosx. 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 (2sinxcosx · e^x)</p>
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<p>Calculate the derivative of (2sinxcosx · e^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>Here, we have f(x) = (2sinxcosx) · e^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 2sinxcosx and v = e^x. Let’s differentiate each term, u′ = d/dx (2sinxcosx) = 2cos(2x) v′ = d/dx (e^x) = e^x Substituting into the given equation, f'(x) = (2cos(2x)) · e^x + (2sinxcosx) · e^x Let’s simplify terms to get the final answer, f'(x) = 2e^x (cos(2x) + sinxcosx) Thus, the derivative of the specified function is 2e^x (cos(2x) + sinxcosx).</p>
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<p>Here, we have f(x) = (2sinxcosx) · e^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 2sinxcosx and v = e^x. Let’s differentiate each term, u′ = d/dx (2sinxcosx) = 2cos(2x) v′ = d/dx (e^x) = e^x Substituting into the given equation, f'(x) = (2cos(2x)) · e^x + (2sinxcosx) · e^x Let’s simplify terms to get the final answer, f'(x) = 2e^x (cos(2x) + sinxcosx) Thus, the derivative of the specified function is 2e^x (cos(2x) + sinxcosx).</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. The first step is finding its derivative and then combining them using the product rule to get the final result.</p>
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<p>We find the derivative of the given function by dividing it into two parts. The first step is finding its derivative 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 company constructs a bridge with an arch modeled by y = 2sinxcosx. Determine the slope of the bridge at x = π/6 meters.</p>
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<p>A company constructs a bridge with an arch modeled by y = 2sinxcosx. Determine the slope of the bridge at x = π/6 meters.</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 = 2sinxcosx (slope of the bridge)...(1) Now, we will differentiate the equation (1) Take the derivative of 2sinxcosx: dy/dx = 2cos(2x) Given x = π/6, substitute this into the derivative: dy/dx = 2cos(π/3) = 2(1/2) = 1 Hence, the slope of the bridge at x = π/6 is 1.</p>
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<p>We have y = 2sinxcosx (slope of the bridge)...(1) Now, we will differentiate the equation (1) Take the derivative of 2sinxcosx: dy/dx = 2cos(2x) Given x = π/6, substitute this into the derivative: dy/dx = 2cos(π/3) = 2(1/2) = 1 Hence, the slope of the bridge at x = π/6 is 1.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the slope of the bridge at x = π/6 as 1, which means that at this point, the elevation of the bridge changes at a rate equal to the horizontal distance.</p>
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<p>We find the slope of the bridge at x = π/6 as 1, which means that at this point, the elevation of the bridge changes at a rate equal to the horizontal distance.</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 = 2sinxcosx.</p>
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<p>Derive the second derivative of the function y = 2sinxcosx.</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 = 2cos(2x)...(1) Now, we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2cos(2x)] Using the chain rule, d²y/dx² = -4sin(2x) Therefore, the second derivative of the function y = 2sinxcosx is -4sin(2x).</p>
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<p>The first step is to find the first derivative, dy/dx = 2cos(2x)...(1) Now, we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2cos(2x)] Using the chain rule, d²y/dx² = -4sin(2x) Therefore, the second derivative of the function y = 2sinxcosx is -4sin(2x).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use a step-by-step process, starting with the first derivative. Using the chain rule, we differentiate 2cos(2x), substitute the identity, and simplify the terms to find the final answer.</p>
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<p>We use a step-by-step process, starting with the first derivative. Using the chain rule, we differentiate 2cos(2x), substitute the identity, and simplify the terms to find the final answer.</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 ((2sinxcosx)²) = 8sinxcosxcos(2x).</p>
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<p>Prove: d/dx ((2sinxcosx)²) = 8sinxcosxcos(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 = (2sinxcosx)² [simplifying] y = (sin(2x))² To differentiate, we use the chain rule: dy/dx = 2sin(2x) · d/dx [sin(2x)] Since the derivative of sin(2x) is 2cos(2x), dy/dx = 2sin(2x) · 2cos(2x) = 4sin(2x)cos(2x) Substituting 2sinxcosx = sin(2x), d/dx ((2sinxcosx)²) = 8sinxcosxcos(2x) Hence proved.</p>
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<p>Let’s start using the chain rule: Consider y = (2sinxcosx)² [simplifying] y = (sin(2x))² To differentiate, we use the chain rule: dy/dx = 2sin(2x) · d/dx [sin(2x)] Since the derivative of sin(2x) is 2cos(2x), dy/dx = 2sin(2x) · 2cos(2x) = 4sin(2x)cos(2x) Substituting 2sinxcosx = sin(2x), d/dx ((2sinxcosx)²) = 8sinxcosxcos(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. Then, we replaced sin(2x) with its derivative. As a final step, we substituted back to derive the equation.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replaced sin(2x) with its derivative. As a final step, we substituted back to derive the equation.</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 (2sinxcosx/x)</p>
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<p>Solve: d/dx (2sinxcosx/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 (2sinxcosx/x) = (d/dx (2sinxcosx) · x - 2sinxcosx · d/dx(x)) / x² We substitute d/dx (2sinxcosx) = 2cos(2x) and d/dx (x) = 1 = (2cos(2x) · x - 2sinxcosx · 1) / x² = (2xcos(2x) - 2sinxcosx) / x² Therefore, d/dx (2sinxcosx/x) = (2xcos(2x) - 2sinxcosx) / x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (2sinxcosx/x) = (d/dx (2sinxcosx) · x - 2sinxcosx · d/dx(x)) / x² We substitute d/dx (2sinxcosx) = 2cos(2x) and d/dx (x) = 1 = (2cos(2x) · x - 2sinxcosx · 1) / x² = (2xcos(2x) - 2sinxcosx) / x² Therefore, d/dx (2sinxcosx/x) = (2xcos(2x) - 2sinxcosx) / 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 product rule and quotient rule. As a final step, we simplify the equation to obtain the final result.</p>
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<p>In this process, we differentiate the given function using the product rule and quotient rule. 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 2sinxcosx</h2>
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<h2>FAQs on the Derivative of 2sinxcosx</h2>
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<h3>1.Find the derivative of 2sinxcosx.</h3>
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<h3>1.Find the derivative of 2sinxcosx.</h3>
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<p>Using the double angle identity, 2sinxcosx simplifies to sin(2x). Differentiating gives: d/dx (sin(2x)) = 2cos(2x)</p>
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<p>Using the double angle identity, 2sinxcosx simplifies to sin(2x). Differentiating gives: d/dx (sin(2x)) = 2cos(2x)</p>
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<h3>2.Can we use the derivative of 2sinxcosx in real life?</h3>
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<h3>2.Can we use the derivative of 2sinxcosx in real life?</h3>
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<p>Yes, the derivative of 2sinxcosx can be applied in physics and engineering for analyzing wave functions and oscillations.</p>
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<p>Yes, the derivative of 2sinxcosx can be applied in physics and engineering for analyzing wave functions and oscillations.</p>
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<h3>3.Is it possible to take the derivative of 2sinxcosx at the point where x = π/2?</h3>
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<h3>3.Is it possible to take the derivative of 2sinxcosx at the point where x = π/2?</h3>
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<p>Yes, at x = π/2, the derivative of 2sinxcosx is 2cos(π) = -2.</p>
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<p>Yes, at x = π/2, the derivative of 2sinxcosx is 2cos(π) = -2.</p>
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<h3>4.What rule is used to differentiate 2sinxcosx/x?</h3>
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<h3>4.What rule is used to differentiate 2sinxcosx/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate 2sinxcosx/x: d/dx (2sinxcosx/x) = (2xcos(2x) - 2sinxcosx) / x².</p>
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<p>We use the<a>quotient</a>rule to differentiate 2sinxcosx/x: d/dx (2sinxcosx/x) = (2xcos(2x) - 2sinxcosx) / x².</p>
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<h3>5.Are the derivatives of 2sinxcosx and sin(2x) the same?</h3>
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<h3>5.Are the derivatives of 2sinxcosx and sin(2x) the same?</h3>
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<p>Yes, because 2sinxcosx simplifies to sin(2x), their derivatives are the same: 2cos(2x).</p>
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<p>Yes, because 2sinxcosx simplifies to sin(2x), their derivatives are the same: 2cos(2x).</p>
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<h2>Important Glossaries for the Derivative of 2sinxcosx</h2>
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<h2>Important Glossaries for the Derivative of 2sinxcosx</h2>
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<p>- Double Angle Identity: A trigonometric identity that expresses 2sinxcosx as sin(2x). - Chain Rule: A rule for differentiating composite functions. - Product Rule: A rule used to differentiate products of two functions. - Sine Function: A primary trigonometric function, expressed as sin(x). - Cosine Function: A primary trigonometric function, expressed as cos(x).</p>
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<p>- Double Angle Identity: A trigonometric identity that expresses 2sinxcosx as sin(2x). - Chain Rule: A rule for differentiating composite functions. - Product Rule: A rule used to differentiate products of two functions. - Sine Function: A primary trigonometric function, expressed as sin(x). - Cosine Function: A primary trigonometric function, expressed as cos(x).</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>