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2026-01-01
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<p>Last updated on<strong>September 22, 2025</strong></p>
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<p>Last updated on<strong>September 22, 2025</strong></p>
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<p>We use the derivative of sin(5x), which is 5cos(5x), as a measuring tool for how the sine function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of sin(5x) in detail.</p>
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<p>We use the derivative of sin(5x), which is 5cos(5x), as a measuring tool for how the sine function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of sin(5x) in detail.</p>
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<h2>What is the Derivative of Sin(5x)?</h2>
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<h2>What is the Derivative of Sin(5x)?</h2>
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<p>We now understand the derivative<a>of</a>sin(5x). It is commonly represented as d/dx (sin(5x)) or (sin(5x))', and its value is 5cos(5x).</p>
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<p>We now understand the derivative<a>of</a>sin(5x). It is commonly represented as d/dx (sin(5x)) or (sin(5x))', and its value is 5cos(5x).</p>
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<p>The<a>function</a>sin(5x) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
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<p>The<a>function</a>sin(5x) 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(5x) is a trigonometric function.</p>
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<p><strong>Sine Function:</strong>sin(5x) is a trigonometric function.</p>
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<p><strong>Chain Rule:</strong>Rule for differentiating sin(5x) (since it involves a composite function).</p>
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<p><strong>Chain Rule:</strong>Rule for differentiating sin(5x) (since it involves a composite function).</p>
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<p><strong>Cosine Function:</strong>cos(x) is another primary trigonometric function.</p>
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<p><strong>Cosine Function:</strong>cos(x) is another primary trigonometric function.</p>
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<h2>Derivative of Sin(5x) Formula</h2>
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<h2>Derivative of Sin(5x) Formula</h2>
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<p>The derivative of sin(5x) can be denoted as d/dx (sin(5x)) or (sin(5x))'. The<a>formula</a>we use to differentiate sin(5x) is: d/dx (sin(5x)) = 5cos(5x) (sin(5x))' = 5cos(5x) The formula applies to all x.</p>
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<p>The derivative of sin(5x) can be denoted as d/dx (sin(5x)) or (sin(5x))'. The<a>formula</a>we use to differentiate sin(5x) is: d/dx (sin(5x)) = 5cos(5x) (sin(5x))' = 5cos(5x) The formula applies to all x.</p>
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<h2>Proofs of the Derivative of Sin(5x)</h2>
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<h2>Proofs of the Derivative of Sin(5x)</h2>
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<p>We can derive the derivative of sin(5x) 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 sin(5x) 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 sin(5x) results in 5cos(5x) using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of sin(5x) results in 5cos(5x) 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 sin(5x) can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>The derivative of sin(5x) can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>To find the derivative of sin(5x) using the first principle, we will consider f(x) = sin(5x). 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>To find the derivative of sin(5x) using the first principle, we will consider f(x) = sin(5x). 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) = sin(5x), we write f(x + h) = sin(5(x + h)).</p>
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<p>Given that f(x) = sin(5x), we write f(x + h) = sin(5(x + h)).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [sin(5(x + h)) - sin(5x)] / h = limₕ→₀ [2cos((5x + 5h + 5x)/2)sin(5h/2)] / h = limₕ→₀ [2cos(5x + 5h/2)sin(5h/2)] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [sin(5(x + h)) - sin(5x)] / h = limₕ→₀ [2cos((5x + 5h + 5x)/2)sin(5h/2)] / h = limₕ→₀ [2cos(5x + 5h/2)sin(5h/2)] / h</p>
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<p>Using limit formulas, limₕ→₀ (sin(5h/2))/(5h/2) = 5/2. f'(x) = 5cos(5x) Hence, proved.</p>
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<p>Using limit formulas, limₕ→₀ (sin(5h/2))/(5h/2) = 5/2. f'(x) = 5cos(5x) 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 sin(5x) using the chain rule, We use the formula:</p>
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<p>To prove the differentiation of sin(5x) using the chain rule, We use the formula:</p>
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<p>Let u = 5x, then sin(u) = sin(5x) d/dx (sin(5x)) = cos(5x) · d/dx (5x) = cos(5x) · 5 = 5cos(5x).</p>
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<p>Let u = 5x, then sin(u) = sin(5x) d/dx (sin(5x)) = cos(5x) · d/dx (5x) = cos(5x) · 5 = 5cos(5x).</p>
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<h2>Higher-Order Derivatives of Sin(5x)</h2>
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<h2>Higher-Order Derivatives of Sin(5x)</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.</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.</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 make it easier to understand functions like sin(5x).</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 make it easier to understand functions like sin(5x).</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 sin(5x), we generally use fⁿ(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 sin(5x), we generally use fⁿ(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 π/2, the derivative is 0 because cos(5x) is 0 at x = π/2. When x is 0, the derivative of sin(5x) = 5cos(0), which is 5.</p>
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<p>When x is π/2, the derivative is 0 because cos(5x) is 0 at x = π/2. When x is 0, the derivative of sin(5x) = 5cos(0), which is 5.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Sin(5x)</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Sin(5x)</h2>
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<p>Students frequently make mistakes when differentiating sin(5x). 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 sin(5x). 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 (sin(5x)·cos(5x))</p>
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<p>Calculate the derivative of (sin(5x)·cos(5x))</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) = sin(5x)·cos(5x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin(5x) and v = cos(5x).</p>
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<p>Here, we have f(x) = sin(5x)·cos(5x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin(5x) and v = cos(5x).</p>
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<p>Let’s differentiate each term, u′ = d/dx (sin(5x)) = 5cos(5x) v′ = d/dx (cos(5x)) = -5sin(5x)</p>
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<p>Let’s differentiate each term, u′ = d/dx (sin(5x)) = 5cos(5x) v′ = d/dx (cos(5x)) = -5sin(5x)</p>
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<p>Substituting into the given equation, f'(x) = (5cos(5x))(cos(5x)) + (sin(5x))(-5sin(5x))</p>
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<p>Substituting into the given equation, f'(x) = (5cos(5x))(cos(5x)) + (sin(5x))(-5sin(5x))</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 5cos²(5x) - 5sin²(5x)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 5cos²(5x) - 5sin²(5x)</p>
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<p>Thus, the derivative of the specified function is 5cos²(5x) - 5sin²(5x).</p>
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<p>Thus, the derivative of the specified function is 5cos²(5x) - 5sin²(5x).</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 factory monitors temperature changes using the function y = sin(5x), where y represents the temperature at time x. If x = π/4 minutes, calculate the rate of change of temperature.</p>
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<p>A factory monitors temperature changes using the function y = sin(5x), where y represents the temperature at time x. If x = π/4 minutes, calculate the rate of change of temperature.</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 = sin(5x) (temperature function)...(1)</p>
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<p>We have y = sin(5x) (temperature function)...(1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Take the derivative of sin(5x): dy/dx = 5cos(5x)</p>
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<p>Take the derivative of sin(5x): dy/dx = 5cos(5x)</p>
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<p>Given x = π/4 (substitute this into the derivative)</p>
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<p>Given x = π/4 (substitute this into the derivative)</p>
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<p>dy/dx = 5cos(5(π/4)) = 5cos(5π/4) Since cos(5π/4) = -√2/2, dy/dx = 5(-√2/2) = -5√2/2</p>
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<p>dy/dx = 5cos(5(π/4)) = 5cos(5π/4) Since cos(5π/4) = -√2/2, dy/dx = 5(-√2/2) = -5√2/2</p>
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<p>Hence, the rate of change of temperature at x = π/4 minutes is -5√2/2.</p>
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<p>Hence, the rate of change of temperature at x = π/4 minutes is -5√2/2.</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 temperature at x = π/4 as -5√2/2, which indicates the temperature is decreasing at that specific moment.</p>
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<p>We find the rate of change of temperature at x = π/4 as -5√2/2, which indicates the temperature is decreasing at that specific 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 = sin(5x).</p>
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<p>Derive the second derivative of the function y = sin(5x).</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 = 5cos(5x)...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = 5cos(5x)...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [5cos(5x)] = 5[-5sin(5x)] = -25sin(5x)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [5cos(5x)] = 5[-5sin(5x)] = -25sin(5x)</p>
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<p>Therefore, the second derivative of the function y = sin(5x) is -25sin(5x).</p>
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<p>Therefore, the second derivative of the function y = sin(5x) is -25sin(5x).</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 5cos(5x). 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 5cos(5x). 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 (sin²(5x)) = 10sin(5x)cos(5x).</p>
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<p>Prove: d/dx (sin²(5x)) = 10sin(5x)cos(5x).</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(5x) = [sin(5x)]²</p>
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<p>Let’s start using the chain rule: Consider y = sin²(5x) = [sin(5x)]²</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2sin(5x)·d/dx [sin(5x)]</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2sin(5x)·d/dx [sin(5x)]</p>
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<p>Since the derivative of sin(5x) is 5cos(5x), dy/dx = 2sin(5x)·5cos(5x) = 10sin(5x)cos(5x)</p>
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<p>Since the derivative of sin(5x) is 5cos(5x), dy/dx = 2sin(5x)·5cos(5x) = 10sin(5x)cos(5x)</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 sin(5x) with its derivative. 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 sin(5x) with its derivative. 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 (sin(5x)/x)</p>
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<p>Solve: d/dx (sin(5x)/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 (sin(5x)/x) = (d/dx (sin(5x))⋅x - sin(5x)⋅d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (sin(5x)/x) = (d/dx (sin(5x))⋅x - sin(5x)⋅d/dx(x))/x²</p>
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<p>We will substitute d/dx (sin(5x)) = 5cos(5x) and d/dx (x) = 1 = (5cos(5x)⋅x - sin(5x)⋅1)/x² = (5xcos(5x) - sin(5x))/x²</p>
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<p>We will substitute d/dx (sin(5x)) = 5cos(5x) and d/dx (x) = 1 = (5cos(5x)⋅x - sin(5x)⋅1)/x² = (5xcos(5x) - sin(5x))/x²</p>
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<p>Therefore, d/dx (sin(5x)/x) = (5xcos(5x) - sin(5x))/x²</p>
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<p>Therefore, d/dx (sin(5x)/x) = (5xcos(5x) - sin(5x))/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(5x)</h2>
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<h2>FAQs on the Derivative of Sin(5x)</h2>
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<h3>1.Find the derivative of sin(5x).</h3>
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<h3>1.Find the derivative of sin(5x).</h3>
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<p>Using the chain rule for sin(5x), we have: d/dx (sin(5x)) = 5cos(5x).</p>
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<p>Using the chain rule for sin(5x), we have: d/dx (sin(5x)) = 5cos(5x).</p>
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<h3>2.Can we use the derivative of sin(5x) in real life?</h3>
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<h3>2.Can we use the derivative of sin(5x) in real life?</h3>
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<p>Yes, we can use the derivative of sin(5x) in real life in calculating the rate of change of any motion, especially in fields such as mathematics, physics, and engineering.</p>
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<p>Yes, we can use the derivative of sin(5x) in real life in calculating the rate of change of any motion, especially in fields such as mathematics, physics, and engineering.</p>
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<h3>3.Is it possible to take the derivative of sin(5x) at the point where x = π/2?</h3>
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<h3>3.Is it possible to take the derivative of sin(5x) at the point where x = π/2?</h3>
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<p>Yes, the derivative at x = π/2 is defined and can be calculated as 5cos(5π/2).</p>
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<p>Yes, the derivative at x = π/2 is defined and can be calculated as 5cos(5π/2).</p>
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<h3>4.What rule is used to differentiate sin(5x)/x?</h3>
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<h3>4.What rule is used to differentiate sin(5x)/x?</h3>
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<p>We use the quotient rule to differentiate sin(5x)/x: d/dx (sin(5x)/x) = (x⋅5cos(5x) - sin(5x)⋅1)/x².</p>
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<p>We use the quotient rule to differentiate sin(5x)/x: d/dx (sin(5x)/x) = (x⋅5cos(5x) - sin(5x)⋅1)/x².</p>
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<h3>5.Are the derivatives of sin(5x) and sin⁻¹(5x) the same?</h3>
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<h3>5.Are the derivatives of sin(5x) and sin⁻¹(5x) the same?</h3>
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<p>No, they are different. The derivative of sin(5x) is 5cos(5x), while the derivative of sin⁻¹(x) is 1/√(1-x²).</p>
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<p>No, they are different. The derivative of sin(5x) is 5cos(5x), while the derivative of sin⁻¹(x) is 1/√(1-x²).</p>
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<h3>6.Can we find the derivative of the sin(5x) formula?</h3>
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<h3>6.Can we find the derivative of the sin(5x) formula?</h3>
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<p>To find, consider y = sin(5x). We use the chain rule: y' = cos(5x)·d/dx(5x) = 5cos(5x).</p>
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<p>To find, consider y = sin(5x). We use the chain rule: y' = cos(5x)·d/dx(5x) = 5cos(5x).</p>
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<h2>Important Glossaries for the Derivative of Sin(5x)</h2>
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<h2>Important Glossaries for the Derivative of Sin(5x)</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 six 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 six trigonometric functions and is written as sin(x).</li>
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</ul><ul><li><strong>Cosine Function:</strong>Another primary trigonometric function, written as cos(x), representing the adjacent side over the hypotenuse in a right triangle.</li>
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</ul><ul><li><strong>Cosine Function:</strong>Another primary trigonometric function, written as cos(x), representing the adjacent side over the hypotenuse in a right triangle.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A fundamental rule in calculus used to differentiate composite functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A fundamental rule in calculus used to differentiate composite functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A method for finding the derivative of a function that is the ratio of two differentiable functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A method for finding the derivative of a function that is the ratio of two differentiable 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>