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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 cos(x/3) to understand how the cosine function changes in response to a slight change in x. Derivatives are crucial tools in various fields, including physics and engineering, for analyzing waveforms and oscillations. We will now discuss the derivative of cos(x/3) in detail.</p>
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<p>We use the derivative of cos(x/3) to understand how the cosine function changes in response to a slight change in x. Derivatives are crucial tools in various fields, including physics and engineering, for analyzing waveforms and oscillations. We will now discuss the derivative of cos(x/3) in detail.</p>
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<h2>What is the Derivative of cos(x/3)?</h2>
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<h2>What is the Derivative of cos(x/3)?</h2>
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<p>We will explore the derivative of cos(x/3). It is commonly represented as d/dx [cos(x/3)] or [cos(x/3)]', and its value is -(1/3)sin(x/3). The<a>function</a>cos(x/3) has a well-defined derivative, indicating it is differentiable within its domain.</p>
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<p>We will explore the derivative of cos(x/3). It is commonly represented as d/dx [cos(x/3)] or [cos(x/3)]', and its value is -(1/3)sin(x/3). The<a>function</a>cos(x/3) has a well-defined derivative, indicating it is differentiable within its domain.</p>
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<p>The key concepts are mentioned below:</p>
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<p>The key concepts are mentioned below:</p>
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<p>Cosine Function: cos(x/3) is a cosine function with a horizontal stretch.</p>
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<p>Cosine Function: cos(x/3) is a cosine function with a horizontal stretch.</p>
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<p>Chain Rule: Rule for differentiating composite functions like cos(x/3).</p>
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<p>Chain Rule: Rule for differentiating composite functions like cos(x/3).</p>
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<p>Sine Function: sin(x) is related to the derivative of cos(x).</p>
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<p>Sine Function: sin(x) is related to the derivative of cos(x).</p>
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<h2>Derivative of cos(x/3) Formula</h2>
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<h2>Derivative of cos(x/3) Formula</h2>
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<p>The derivative of cos(x/3) can be denoted as d/dx [cos(x/3)] or [cos(x/3)]'.</p>
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<p>The derivative of cos(x/3) can be denoted as d/dx [cos(x/3)] or [cos(x/3)]'.</p>
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<p>The<a>formula</a>we use to differentiate cos(x/3) is: d/dx [cos(x/3)] = -(1/3)sin(x/3)</p>
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<p>The<a>formula</a>we use to differentiate cos(x/3) is: d/dx [cos(x/3)] = -(1/3)sin(x/3)</p>
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<p>The formula applies to all x where cos(x/3) is defined.</p>
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<p>The formula applies to all x where cos(x/3) is defined.</p>
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<h2>Proofs of the Derivative of cos(x/3)</h2>
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<h2>Proofs of the Derivative of cos(x/3)</h2>
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<p>We can derive the derivative of cos(x/3) using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of cos(x/3) using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
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<ol><li>By First Principle</li>
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<ol><li>By First Principle</li>
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<li>Using Chain Rule</li>
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<li>Using Chain Rule</li>
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</ol><p>We will now demonstrate that the differentiation of cos(x/3) results in -(1/3)sin(x/3) using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of cos(x/3) results in -(1/3)sin(x/3) using the above-mentioned methods:</p>
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<h3>By First Principle</h3>
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<h3>By First Principle</h3>
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<p>The derivative of cos(x/3) can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of cos(x/3) u</p>
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<p>The derivative of cos(x/3) can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of cos(x/3) u</p>
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<p>sing the first principle, we will consider f(x) = cos(x/3). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>sing the first principle, we will consider f(x) = cos(x/3). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>Given that f(x) = cos(x/3), we write f(x + h) = cos((x + h)/3).</p>
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<p>Given that f(x) = cos(x/3), we write f(x + h) = cos((x + h)/3).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [cos((x + h)/3) - cos(x/3)] / h = limₕ→₀ [-2sin((2x + h)/6)sin(h/6)] / h = limₕ→₀ [-sin((2x + h)/6)sin(h/6)] / (3h/6)</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [cos((x + h)/3) - cos(x/3)] / h = limₕ→₀ [-2sin((2x + h)/6)sin(h/6)] / h = limₕ→₀ [-sin((2x + h)/6)sin(h/6)] / (3h/6)</p>
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<p>Using limit formulas, as h approaches zero, sin(h/6)/(h/6) = 1. f'(x) = [-1/3]sin(x/3)</p>
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<p>Using limit formulas, as h approaches zero, sin(h/6)/(h/6) = 1. f'(x) = [-1/3]sin(x/3)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</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 prove the differentiation of cos(x/3) using the chain rule, We use the formula: Cos(x/3) is a composition<a>of functions</a>. Consider u = x/3, then f(u) = cos(u)</p>
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<p>To prove the differentiation of cos(x/3) using the chain rule, We use the formula: Cos(x/3) is a composition<a>of functions</a>. Consider u = x/3, then f(u) = cos(u)</p>
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<p>The derivative of f(u) is -sin(u) and the derivative of u is 1/3.</p>
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<p>The derivative of f(u) is -sin(u) and the derivative of u is 1/3.</p>
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<p>Using the chain rule: d/dx [cos(u)] = -sin(u) * du/dx. So we get, d/dx [cos(x/3)] = -sin(x/3) * (1/3) d/dx [cos(x/3)] = -(1/3)sin(x/3)</p>
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<p>Using the chain rule: d/dx [cos(u)] = -sin(u) * du/dx. So we get, d/dx [cos(x/3)] = -sin(x/3) * (1/3) d/dx [cos(x/3)] = -(1/3)sin(x/3)</p>
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<h2>Higher-Order Derivatives of cos(x/3)</h2>
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<h2>Higher-Order Derivatives of cos(x/3)</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 tricky. T</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 tricky. T</p>
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<p>o understand them better, think of a pendulum where the angular displacement changes (first derivative) and the<a>rate</a>at which this changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like cos(x/3).</p>
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<p>o understand them better, think of a pendulum where the angular displacement changes (first derivative) and the<a>rate</a>at which this changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like cos(x/3).</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 cos(x/3), we generally use f n(x) for the nth derivative of a function f(x) which tells us the change in the rate of change. (continuing for higher-order derivatives).</p>
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<p>For the nth Derivative of cos(x/3), we generally use f n(x) for the nth derivative of a function f(x) which tells us the change in the rate of change. (continuing for higher-order derivatives).</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 3π/2, the derivative is zero because sin(x/3) is zero there. When x is 0, the derivative of cos(x/3) = -(1/3)sin(0), which is 0.</p>
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<p>When x is 3π/2, the derivative is zero because sin(x/3) is zero there. When x is 0, the derivative of cos(x/3) = -(1/3)sin(0), which is 0.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of cos(x/3)</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of cos(x/3)</h2>
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<p>Students frequently make mistakes when differentiating cos(x/3). 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 cos(x/3). 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 [cos(x/3)·sin(x/2)]</p>
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<p>Calculate the derivative of [cos(x/3)·sin(x/2)]</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Here, we have f(x) = cos(x/3)·sin(x/2).</p>
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<p>Here, we have f(x) = cos(x/3)·sin(x/2).</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = cos(x/3) and v = sin(x/2).</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = cos(x/3) and v = sin(x/2).</p>
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<p>Let’s differentiate each term, u′ = d/dx [cos(x/3)] = -(1/3)sin(x/3) v′ = d/dx [sin(x/2)] = (1/2)cos(x/2)</p>
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<p>Let’s differentiate each term, u′ = d/dx [cos(x/3)] = -(1/3)sin(x/3) v′ = d/dx [sin(x/2)] = (1/2)cos(x/2)</p>
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<p>Substituting into the given equation, f'(x) = [-(1/3)sin(x/3)]·[sin(x/2)] + [cos(x/3)]·[(1/2)cos(x/2)]</p>
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<p>Substituting into the given equation, f'(x) = [-(1/3)sin(x/3)]·[sin(x/2)] + [cos(x/3)]·[(1/2)cos(x/2)]</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -(1/3)sin(x/3)sin(x/2) + (1/2)cos(x/3)cos(x/2)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -(1/3)sin(x/3)sin(x/2) + (1/2)cos(x/3)cos(x/2)</p>
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<p>Thus, the derivative of the specified function is -(1/3)sin(x/3)sin(x/2) + (1/2)cos(x/3)cos(x/2).</p>
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<p>Thus, the derivative of the specified function is -(1/3)sin(x/3)sin(x/2) + (1/2)cos(x/3)cos(x/2).</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>A rotating wheel has its angular position given by θ = cos(t/3), where θ is the angular position in radians and t is the time in seconds. Find the rate of change of angular position when t = π seconds.</p>
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<p>A rotating wheel has its angular position given by θ = cos(t/3), where θ is the angular position in radians and t is the time in seconds. Find the rate of change of angular position when t = π 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 = cos(t/3) (angular position)...(1)</p>
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<p>We have θ = cos(t/3) (angular position)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative of cos(t/3): dθ/dt = -(1/3)sin(t/3)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative of cos(t/3): dθ/dt = -(1/3)sin(t/3)</p>
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<p>Given t = π (substitute this into the derivative) dθ/dt = -(1/3)sin(π/3) = -(1/3)(√3/2) = -√3/6</p>
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<p>Given t = π (substitute this into the derivative) dθ/dt = -(1/3)sin(π/3) = -(1/3)(√3/2) = -√3/6</p>
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<p>Hence, the rate of change of angular position at t = π seconds is -√3/6 radians per second.</p>
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<p>Hence, the rate of change of angular position at t = π seconds is -√3/6 radians per second.</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 angular position at t = π as -√3/6, which indicates how the angular position is changing at that moment.</p>
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<p>We find the rate of change of angular position at t = π as -√3/6, which indicates how the angular position is changing at that moment.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function y = cos(x/3).</p>
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<p>Derive the second derivative of the function y = cos(x/3).</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 = -(1/3)sin(x/3)...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = -(1/3)sin(x/3)...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-(1/3)sin(x/3)]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-(1/3)sin(x/3)]</p>
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<p>Here we use the chain rule, d²y/dx² = -(1/3)[(1/3)cos(x/3)] = -(1/9)cos(x/3)</p>
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<p>Here we use the chain rule, d²y/dx² = -(1/3)[(1/3)cos(x/3)] = -(1/9)cos(x/3)</p>
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<p>Therefore, the second derivative of the function y = cos(x/3) is -(1/9)cos(x/3).</p>
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<p>Therefore, the second derivative of the function y = cos(x/3) is -(1/9)cos(x/3).</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 chain rule, we differentiate -(1/3)sin(x/3). 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 chain rule, we differentiate -(1/3)sin(x/3). 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 [cos²(x/3)] = -(2/3)cos(x/3)sin(x/3).</p>
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<p>Prove: d/dx [cos²(x/3)] = -(2/3)cos(x/3)sin(x/3).</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 = cos²(x/3) = [cos(x/3)]²</p>
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<p>Let’s start using the chain rule: Consider y = cos²(x/3) = [cos(x/3)]²</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2cos(x/3)·d/dx [cos(x/3)]</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2cos(x/3)·d/dx [cos(x/3)]</p>
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<p>Since the derivative of cos(x/3) is -(1/3)sin(x/3), dy/dx = 2cos(x/3)·[-(1/3)sin(x/3)] = -(2/3)cos(x/3)sin(x/3)</p>
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<p>Since the derivative of cos(x/3) is -(1/3)sin(x/3), dy/dx = 2cos(x/3)·[-(1/3)sin(x/3)] = -(2/3)cos(x/3)sin(x/3)</p>
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<p>Hence proved.</p>
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<p>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 the derivative of cos(x/3) with its expression. As a final step, we simplify 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 replace the derivative of cos(x/3) with its expression. As a final step, we simplify 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 [cos(x/3)/x]</p>
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<p>Solve: d/dx [cos(x/3)/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 [cos(x/3)/x] = (d/dx [cos(x/3)]·x - cos(x/3)·d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx [cos(x/3)/x] = (d/dx [cos(x/3)]·x - cos(x/3)·d/dx(x))/x²</p>
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<p>We will substitute d/dx [cos(x/3)] = -(1/3)sin(x/3) and d/dx(x) = 1 = [-(1/3)sin(x/3)·x - cos(x/3)·1]/x² = [-x(1/3)sin(x/3) - cos(x/3)]/x² = [-(1/3)xsin(x/3) - cos(x/3)]/x²</p>
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<p>We will substitute d/dx [cos(x/3)] = -(1/3)sin(x/3) and d/dx(x) = 1 = [-(1/3)sin(x/3)·x - cos(x/3)·1]/x² = [-x(1/3)sin(x/3) - cos(x/3)]/x² = [-(1/3)xsin(x/3) - cos(x/3)]/x²</p>
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<p>Therefore, d/dx [cos(x/3)/x] = [-(1/3)xsin(x/3) - cos(x/3)]/x²</p>
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<p>Therefore, d/dx [cos(x/3)/x] = [-(1/3)xsin(x/3) - cos(x/3)]/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. 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 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 cos(x/3)</h2>
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<h2>FAQs on the Derivative of cos(x/3)</h2>
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<h3>1.Find the derivative of cos(x/3).</h3>
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<h3>1.Find the derivative of cos(x/3).</h3>
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<p>Using the chain rule for cos(x/3), d/dx [cos(x/3)] = -(1/3)sin(x/3).</p>
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<p>Using the chain rule for cos(x/3), d/dx [cos(x/3)] = -(1/3)sin(x/3).</p>
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<h3>2.Can we use the derivative of cos(x/3) in real life?</h3>
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<h3>2.Can we use the derivative of cos(x/3) in real life?</h3>
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<p>Yes, we can use the derivative of cos(x/3) in real life in analyzing waveforms, oscillations, and other periodic phenomena in various fields like physics and engineering.</p>
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<p>Yes, we can use the derivative of cos(x/3) in real life in analyzing waveforms, oscillations, and other periodic phenomena in various fields like physics and engineering.</p>
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<h3>3.Is it possible to take the derivative of cos(x/3) at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of cos(x/3) at the point where x = 0?</h3>
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<p>Yes, at x = 0, the derivative of cos(x/3) is -(1/3)sin(0) = 0.</p>
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<p>Yes, at x = 0, the derivative of cos(x/3) is -(1/3)sin(0) = 0.</p>
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<h3>4.What rule is used to differentiate cos(x/3)/x?</h3>
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<h3>4.What rule is used to differentiate cos(x/3)/x?</h3>
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<p>We use the quotient rule to differentiate cos(x/3)/x, d/dx [cos(x/3)/x] = [-(1/3)xsin(x/3) - cos(x/3)]/x².</p>
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<p>We use the quotient rule to differentiate cos(x/3)/x, d/dx [cos(x/3)/x] = [-(1/3)xsin(x/3) - cos(x/3)]/x².</p>
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<h3>5.Are the derivatives of cos(x/3) and cos⁻¹(x/3) the same?</h3>
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<h3>5.Are the derivatives of cos(x/3) and cos⁻¹(x/3) the same?</h3>
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<p>No, they are different. The derivative of cos(x/3) is -(1/3)sin(x/3), while the derivative of cos⁻¹(x/3) involves an inverse trigonometric function and is more complex.</p>
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<p>No, they are different. The derivative of cos(x/3) is -(1/3)sin(x/3), while the derivative of cos⁻¹(x/3) involves an inverse trigonometric function and is more complex.</p>
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<h2>Important Glossaries for the Derivative of cos(x/3)</h2>
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<h2>Important Glossaries for the Derivative of cos(x/3)</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>Cosine Function:</strong>A trigonometric function representing the cosine of an angle, often used in waveforms and oscillations.</li>
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</ul><ul><li><strong>Cosine Function:</strong>A trigonometric function representing the cosine of an angle, often used in waveforms and oscillations.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule in calculus for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule in calculus for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Sine Function:</strong>A trigonometric function related to the derivative of the cosine function.</li>
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</ul><ul><li><strong>Sine Function:</strong>A trigonometric function related to the derivative of the cosine function.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A formula for differentiating the division of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A formula for differentiating the division 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>