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2026-01-01
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>We use the derivative of sin³(x), which helps us understand how the function changes in response to a small change in x. Derivatives are useful for calculating various changes in real-life situations. We will now discuss the derivative of sin³(x) in detail.</p>
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<p>We use the derivative of sin³(x), which helps us understand how the function changes in response to a small change in x. Derivatives are useful for calculating various changes in real-life situations. We will now discuss the derivative of sin³(x) in detail.</p>
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<h2>What is the Derivative of Sin Cubed x?</h2>
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<h2>What is the Derivative of Sin Cubed x?</h2>
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<p>We now understand the derivative of sin³(x). It is commonly represented as d/dx (sin³x) or (sin³x)', and its value is 3sin²x·cosx. The<a>function</a>sin³x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
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<p>We now understand the derivative of sin³(x). It is commonly represented as d/dx (sin³x) or (sin³x)', and its value is 3sin²x·cosx. The<a>function</a>sin³x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
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<p><strong>Sine Function:</strong>(sin(x)).</p>
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<p><strong>Sine Function:</strong>(sin(x)).</p>
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<p><strong>Chain Rule:</strong>Rule for differentiating composite functions like sin³x.</p>
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<p><strong>Chain Rule:</strong>Rule for differentiating composite functions like sin³x.</p>
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<p><strong>Power Rule:</strong>Rule for differentiating functions of the form x^n.</p>
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<p><strong>Power Rule:</strong>Rule for differentiating functions of the form x^n.</p>
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<h2>Derivative of Sin Cubed x Formula</h2>
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<h2>Derivative of Sin Cubed x Formula</h2>
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<p>The derivative of sin³x can be denoted as d/dx (sin³x) or (sin³x)'.</p>
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<p>The derivative of sin³x can be denoted as d/dx (sin³x) or (sin³x)'.</p>
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<p>The<a>formula</a>we use to differentiate sin³x is: d/dx (sin³x) = 3sin²x·cosx</p>
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<p>The<a>formula</a>we use to differentiate sin³x is: d/dx (sin³x) = 3sin²x·cosx</p>
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<p>The formula applies to all x where sin(x) is defined.</p>
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<p>The formula applies to all x where sin(x) is defined.</p>
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<h2>Proofs of the Derivative of Sin Cubed x</h2>
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<h2>Proofs of the Derivative of Sin Cubed x</h2>
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<p>We can derive the derivative of sin³x using proofs. To show this, we will use trigonometric identities along with differentiation rules. There are several methods we use to prove this, such as: Using the Chain Rule Using the Product Rule We will now demonstrate that the differentiation of sin³x results in 3sin²x·cosx using these methods:</p>
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<p>We can derive the derivative of sin³x using proofs. To show this, we will use trigonometric identities along with differentiation rules. There are several methods we use to prove this, such as: Using the Chain Rule Using the Product Rule We will now demonstrate that the differentiation of sin³x results in 3sin²x·cosx using these methods:</p>
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<h3>Using the Chain Rule</h3>
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<h3>Using the Chain Rule</h3>
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<p>To prove the differentiation of sin³x using the chain rule, We use the formula: Let y = (sin(x))³</p>
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<p>To prove the differentiation of sin³x using the chain rule, We use the formula: Let y = (sin(x))³</p>
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<p>By the chain rule, dy/dx = 3(sin(x))²·d/dx(sin(x))</p>
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<p>By the chain rule, dy/dx = 3(sin(x))²·d/dx(sin(x))</p>
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<p>Since d/dx(sin(x)) = cos(x), dy/dx = 3(sin(x))²·cos(x) Hence, proved.</p>
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<p>Since d/dx(sin(x)) = cos(x), dy/dx = 3(sin(x))²·cos(x) Hence, proved.</p>
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<h3>Using the Product Rule</h3>
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<h3>Using the Product Rule</h3>
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<p>We can also prove the derivative of sin³x using the<a>product</a>rule. Here's the step-by-step process: Let y = sin(x)·sin(x)·sin(x)</p>
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<p>We can also prove the derivative of sin³x using the<a>product</a>rule. Here's the step-by-step process: Let y = sin(x)·sin(x)·sin(x)</p>
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<p>By the product rule, d/dx[u·v·w] = uv'w + u'vw + uvw' Let u = sin(x), v = sin(x), w = sin(x)</p>
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<p>By the product rule, d/dx[u·v·w] = uv'w + u'vw + uvw' Let u = sin(x), v = sin(x), w = sin(x)</p>
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<p>Then u' = cos(x), v' = cos(x), w' = cos(x) dy/dx = sin(x)·sin(x)·cos(x) + sin(x)·cos(x)·sin(x) + cos(x)·sin(x)·sin(x) = 3sin²(x)·cos(x)</p>
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<p>Then u' = cos(x), v' = cos(x), w' = cos(x) dy/dx = sin(x)·sin(x)·cos(x) + sin(x)·cos(x)·sin(x) + cos(x)·sin(x)·sin(x) = 3sin²(x)·cos(x)</p>
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<p>Thus, d/dx(sin³x) = 3sin²x·cosx.</p>
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<p>Thus, d/dx(sin³x) = 3sin²x·cosx.</p>
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<h2>Higher-Order Derivatives of Sin Cubed x</h2>
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<h2>Higher-Order Derivatives of Sin Cubed x</h2>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little complex.</p>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little complex.</p>
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<p>To understand them better, think of 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 help in understanding functions like sin³(x).</p>
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<p>To understand them better, think of 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 help in understanding functions like sin³(x).</p>
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<p>For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>For the nth Derivative of sin³(x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.</p>
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<p>For the nth Derivative of sin³(x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us 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>When x is 0, the derivative of sin³(x) = 3sin²(0)·cos(0), which is 0.</p>
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<p>When x is 0, the derivative of sin³(x) = 3sin²(0)·cos(0), which is 0.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Sin Cubed x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Sin Cubed x</h2>
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<p>Students frequently make mistakes when differentiating sin³x. 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 sin³x. 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 (sin³x·cos²x)</p>
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<p>Calculate the derivative of (sin³x·cos²x)</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) = sin³x·cos²x.</p>
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<p>Here, we have f(x) = sin³x·cos²x.</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin³x and v = cos²x.</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin³x and v = cos²x.</p>
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<p>Let’s differentiate each term, u′= d/dx (sin³x) = 3sin²x·cosx v′= d/dx (cos²x) = -2cosx·sinx</p>
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<p>Let’s differentiate each term, u′= d/dx (sin³x) = 3sin²x·cosx v′= d/dx (cos²x) = -2cosx·sinx</p>
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<p>Substituting into the given equation, f'(x) = (3sin²x·cosx)·(cos²x) + (sin³x)·(-2cosx·sinx)</p>
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<p>Substituting into the given equation, f'(x) = (3sin²x·cosx)·(cos²x) + (sin³x)·(-2cosx·sinx)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 3sin²x·cos³x - 2sin⁴x·cosx</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 3sin²x·cos³x - 2sin⁴x·cosx</p>
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<p>Thus, the derivative of the specified function is 3sin²x·cos³x - 2sin⁴x·cosx.</p>
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<p>Thus, the derivative of the specified function is 3sin²x·cos³x - 2sin⁴x·cosx.</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 the function 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 the function 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>The height of a tree is modeled by the function y = sin³(x) meters, where x is time in months. Calculate the rate of change of the tree's height when x = π/6 months.</p>
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<p>The height of a tree is modeled by the function y = sin³(x) meters, where x is time in months. Calculate the rate of change of the tree's height when x = π/6 months.</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 = sin³(x) (height of the tree)...(1)</p>
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<p>We have y = sin³(x) (height of the tree)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative of sin³(x): dy/dx = 3sin²x·cosx</p>
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<p>Now, we will differentiate the equation (1) Take the derivative of sin³(x): dy/dx = 3sin²x·cosx</p>
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<p>Given x = π/6, substitute this into the derivative: dy/dx = 3sin²(π/6)·cos(π/6) = 3(1/2)²·(√3/2) = 3(1/4)·(√3/2) = 3√3/8</p>
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<p>Given x = π/6, substitute this into the derivative: dy/dx = 3sin²(π/6)·cos(π/6) = 3(1/2)²·(√3/2) = 3(1/4)·(√3/2) = 3√3/8</p>
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<p>Hence, the rate of change of the tree's height at x = π/6 is 3√3/8 meters per month.</p>
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<p>Hence, the rate of change of the tree's height at x = π/6 is 3√3/8 meters per month.</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 tree's height at x = π/6 by differentiating the function y = sin³(x) and then substituting x = π/6 into the derivative.</p>
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<p>We find the rate of change of the tree's height at x = π/6 by differentiating the function y = sin³(x) and then substituting x = π/6 into the derivative.</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 = sin³(x).</p>
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<p>Derive the second derivative of the function y = sin³(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>The first step is to find the first derivative, dy/dx = 3sin²(x)·cos(x)...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = 3sin²(x)·cos(x)...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [3sin²(x)·cos(x)]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [3sin²(x)·cos(x)]</p>
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<p>Here we use the product rule, d²y/dx² = 3[d/dx(sin²(x))·cos(x) + sin²(x)·d/dx(cos(x))] = 3[2sin(x)cos(x)·cos(x) - sin²(x)sin(x)] = 6sin(x)cos²(x) - 3sin³(x)</p>
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<p>Here we use the product rule, d²y/dx² = 3[d/dx(sin²(x))·cos(x) + sin²(x)·d/dx(cos(x))] = 3[2sin(x)cos(x)·cos(x) - sin²(x)sin(x)] = 6sin(x)cos²(x) - 3sin³(x)</p>
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<p>Therefore, the second derivative of the function y = sin³(x) is 6sin(x)cos²(x) - 3sin³(x).</p>
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<p>Therefore, the second derivative of the function y = sin³(x) is 6sin(x)cos²(x) - 3sin³(x).</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, where we start with the first derivative. Using the product rule, we differentiate 3sin²(x)·cos(x). We then simplify the terms to find the final answer.</p>
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<p>We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate 3sin²(x)·cos(x). We then 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 ((sin(x))³) = 3sin²(x)cos(x).</p>
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<p>Prove: d/dx ((sin(x))³) = 3sin²(x)cos(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>Let’s start using the chain rule: Consider y = (sin(x))³</p>
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<p>Let’s start using the chain rule: Consider y = (sin(x))³</p>
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<p>To differentiate, we use the chain rule: dy/dx = 3(sin(x))²·d/dx[sin(x)]</p>
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<p>To differentiate, we use the chain rule: dy/dx = 3(sin(x))²·d/dx[sin(x)]</p>
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<p>Since the derivative of sin(x) is cos(x), dy/dx = 3(sin(x))²·cos(x) Hence proved.</p>
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<p>Since the derivative of sin(x) is cos(x), dy/dx = 3(sin(x))²·cos(x) 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 replace sin(x) with its derivative. As a final step, we obtain the derivative of the function.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sin(x) with its derivative. As a final step, we obtain the derivative of the function.</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 (sin³x/x)</p>
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<p>Solve: d/dx (sin³x/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 (sin³x/x) = (d/dx (sin³x)·x - sin³x·d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (sin³x/x) = (d/dx (sin³x)·x - sin³x·d/dx(x))/x²</p>
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<p>We will substitute d/dx(sin³x) = 3sin²x·cosx and d/dx(x) = 1 = (3sin²x·cosx·x - sin³x·1) / x² = (3xsin²x·cosx - sin³x) / x²</p>
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<p>We will substitute d/dx(sin³x) = 3sin²x·cosx and d/dx(x) = 1 = (3sin²x·cosx·x - sin³x·1) / x² = (3xsin²x·cosx - sin³x) / x²</p>
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<p>Therefore, d/dx (sin³x/x) = (3xsin²x·cosx - sin³x) / x²</p>
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<p>Therefore, d/dx (sin³x/x) = (3xsin²x·cosx - sin³x) / 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 Sin Cubed x</h2>
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<h2>FAQs on the Derivative of Sin Cubed x</h2>
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<h3>1.Find the derivative of sin³x.</h3>
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<h3>1.Find the derivative of sin³x.</h3>
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<p>Using the chain rule on sin³x gives us: d/dx (sin³x) = 3sin²x·cosx (simplified)</p>
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<p>Using the chain rule on sin³x gives us: d/dx (sin³x) = 3sin²x·cosx (simplified)</p>
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<h3>2.Can we use the derivative of sin³x in real life?</h3>
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<h3>2.Can we use the derivative of sin³x in real life?</h3>
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<p>Yes, we can use the derivative of sin³x in real life in calculating rates of change in oscillatory motions, especially in fields such as physics and engineering.</p>
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<p>Yes, we can use the derivative of sin³x in real life in calculating rates of change in oscillatory motions, especially in fields such as physics and engineering.</p>
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<h3>3.Is it possible to take the derivative of sin³x at x = π/2?</h3>
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<h3>3.Is it possible to take the derivative of sin³x at x = π/2?</h3>
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<p>Yes, the derivative at x = π/2 is possible since sin³x is defined at x = π/2, and the derivative is 3sin²(π/2)·cos(π/2) = 0.</p>
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<p>Yes, the derivative at x = π/2 is possible since sin³x is defined at x = π/2, and the derivative is 3sin²(π/2)·cos(π/2) = 0.</p>
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<h3>4.What rule is used to differentiate sin³x/x?</h3>
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<h3>4.What rule is used to differentiate sin³x/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate sin³x/x: d/dx (sin³x/x) = (x·3sin²x·cosx - sin³x)/x².</p>
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<p>We use the<a>quotient</a>rule to differentiate sin³x/x: d/dx (sin³x/x) = (x·3sin²x·cosx - sin³x)/x².</p>
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<h3>5.Are the derivatives of sin³x and (sinx)⁻¹ the same?</h3>
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<h3>5.Are the derivatives of sin³x and (sinx)⁻¹ the same?</h3>
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<p>No, they are different. The derivative of sin³x is 3sin²x·cosx, while the derivative of (sinx)⁻¹ is -cosx/sin²x.</p>
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<p>No, they are different. The derivative of sin³x is 3sin²x·cosx, while the derivative of (sinx)⁻¹ is -cosx/sin²x.</p>
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<h2>Important Glossaries for the Derivative of Sin Cubed x</h2>
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<h2>Important Glossaries for the Derivative of Sin Cubed x</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
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</ul><ul><li><strong>Sine Function:</strong>The sine function is one of the primary trigonometric functions and is written as sin(x).</li>
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</ul><ul><li><strong>Sine Function:</strong>The sine function is one of the primary trigonometric functions and is written as sin(x).</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating composite functions, where you differentiate the outer function and multiply by the derivative of the inner function.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating composite functions, where you differentiate the outer function and multiply by the derivative of the inner function.</li>
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</ul><ul><li><strong>Power Rule:</strong>A rule for differentiating functions of the form x^n, which states that the derivative is n*x^(n-1).</li>
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</ul><ul><li><strong>Power Rule:</strong>A rule for differentiating functions of the form x^n, which states that the derivative is n*x^(n-1).</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating the quotient of two functions, where the derivative is given by (v·u' - u·v')/v².</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating the quotient of two functions, where the derivative is given by (v·u' - u·v')/v².</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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<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>