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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 5sin(x), which is 5cos(x), 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 5sin(x) in detail.</p>
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<p>We use the derivative of 5sin(x), which is 5cos(x), 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 5sin(x) in detail.</p>
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<h2>What is the Derivative of 5sinx?</h2>
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<h2>What is the Derivative of 5sinx?</h2>
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<p>We now understand the derivative of 5sin(x). It is commonly represented as d/dx (5sinx) or (5sinx)', and its value is 5cos(x). </p>
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<p>We now understand the derivative of 5sin(x). It is commonly represented as d/dx (5sinx) or (5sinx)', and its value is 5cos(x). </p>
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<p>The<a>function</a>5sin(x) has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>The<a>function</a>5sin(x) has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>The key concepts are mentioned below: - Sine Function: sin(x) is a trigonometric function.</p>
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<p>The key concepts are mentioned below: - Sine Function: sin(x) is a trigonometric function.</p>
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<p>- Constant Multiple Rule: Rule for differentiating 5sin(x) (since it involves a<a>constant</a><a>multiple</a>of sin(x)).</p>
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<p>- Constant Multiple Rule: Rule for differentiating 5sin(x) (since it involves a<a>constant</a><a>multiple</a>of sin(x)).</p>
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<p>- Cosine Function: cos(x) is the derivative of sin(x).</p>
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<p>- Cosine Function: cos(x) is the derivative of sin(x).</p>
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<h2>Derivative of 5sinx Formula</h2>
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<h2>Derivative of 5sinx Formula</h2>
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<p>The derivative of 5sin(x) can be denoted as d/dx (5sinx) or (5sinx)'. </p>
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<p>The derivative of 5sin(x) can be denoted as d/dx (5sinx) or (5sinx)'. </p>
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<p>The<a>formula</a>we use to differentiate 5sin(x) is: d/dx (5sinx) = 5cos(x) (5sinx)' = 5cos(x) The formula applies to all x.</p>
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<p>The<a>formula</a>we use to differentiate 5sin(x) is: d/dx (5sinx) = 5cos(x) (5sinx)' = 5cos(x) The formula applies to all x.</p>
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<h2>Proofs of the Derivative of 5sinx</h2>
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<h2>Proofs of the Derivative of 5sinx</h2>
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<p>We can derive the derivative of 5sin(x) using proofs. To show this, we will use 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 5sin(x) using proofs. To show this, we will use 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>- By First Principle - Using Constant Multiple Rule We will now demonstrate that the differentiation of 5sin(x) results in 5cos(x) using the above-mentioned methods:</p>
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<p>- By First Principle - Using Constant Multiple Rule We will now demonstrate that the differentiation of 5sin(x) results in 5cos(x) using the above-mentioned methods:</p>
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<p>By First Principle The derivative of 5sin(x) 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>By First Principle The derivative of 5sin(x) 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 5sin(x) using the first principle, we will consider f(x) = 5sin(x).</p>
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<p>To find the derivative of 5sin(x) using the first principle, we will consider f(x) = 5sin(x).</p>
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<p>Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 5sin(x), we write f(x + h) = 5sin(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [5sin(x + h) - 5sin(x)] / h = 5 limₕ→₀ [sin(x + h) - sin(x)] / h We use the formula sin A - sin B = 2cos((A+B)/2)sin((A-B)/2). = 5 limₕ→₀ [2cos((2x + h)/2)sin(h/2)] / h = 5 limₕ→₀ [cos((2x + h)/2)] . limₕ→₀ [sin(h/2)/(h/2)]</p>
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<p>Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 5sin(x), we write f(x + h) = 5sin(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [5sin(x + h) - 5sin(x)] / h = 5 limₕ→₀ [sin(x + h) - sin(x)] / h We use the formula sin A - sin B = 2cos((A+B)/2)sin((A-B)/2). = 5 limₕ→₀ [2cos((2x + h)/2)sin(h/2)] / h = 5 limₕ→₀ [cos((2x + h)/2)] . limₕ→₀ [sin(h/2)/(h/2)]</p>
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<p>Using limit formulas, limₕ→₀ [sin(h)/h] = 1. f'(x) = 5cos(x) Hence, proved.</p>
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<p>Using limit formulas, limₕ→₀ [sin(h)/h] = 1. f'(x) = 5cos(x) Hence, proved.</p>
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<p>Using Constant Multiple Rule To prove the differentiation of 5sin(x) using the constant multiple rule, We use the formula: d/dx [c·f(x)] = c·d/dx [f(x)] Let c = 5 and f(x) = sin(x) d/dx (5sinx) = 5·d/dx (sinx) = 5cos(x)</p>
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<p>Using Constant Multiple Rule To prove the differentiation of 5sin(x) using the constant multiple rule, We use the formula: d/dx [c·f(x)] = c·d/dx [f(x)] Let c = 5 and f(x) = sin(x) d/dx (5sinx) = 5·d/dx (sinx) = 5cos(x)</p>
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<p>Thus, the derivative is 5cos(x).</p>
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<p>Thus, the derivative is 5cos(x).</p>
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<h2>Higher-Order Derivatives of 5sinx</h2>
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<h2>Higher-Order Derivatives of 5sinx</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.</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.</p>
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<p>Higher-order derivatives make it easier to understand functions like 5sin(x). For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point.</p>
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<p>Higher-order derivatives make it easier to understand functions like 5sin(x). For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point.</p>
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<p>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>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 5sin(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 5sin(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 π/2, the derivative is 5cos(π/2), which is 0 because cos(π/2) = 0. When x is 0, the derivative of 5sin(x) = 5cos(0), which is 5 because cos(0) = 1.</p>
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<p>When x is π/2, the derivative is 5cos(π/2), which is 0 because cos(π/2) = 0. When x is 0, the derivative of 5sin(x) = 5cos(0), which is 5 because cos(0) = 1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 5sinx</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 5sinx</h2>
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<p>Students frequently make mistakes when differentiating 5sin(x). 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 5sin(x). 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 (5sinx·cos(x))</p>
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<p>Calculate the derivative of (5sinx·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>Here, we have f(x) = 5sinx·cos(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 5sinx and v = cos(x). </p>
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<p>Here, we have f(x) = 5sinx·cos(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 5sinx and v = cos(x). </p>
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<p>Let’s differentiate each term, u′= d/dx (5sinx) = 5cos(x) v′= d/dx (cos(x)) = -sin(x)</p>
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<p>Let’s differentiate each term, u′= d/dx (5sinx) = 5cos(x) v′= d/dx (cos(x)) = -sin(x)</p>
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<p>Substituting into the given equation, f'(x) = (5cos(x)).(cos(x)) + (5sinx).(-sin(x))</p>
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<p>Substituting into the given equation, f'(x) = (5cos(x)).(cos(x)) + (5sinx).(-sin(x))</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 5cos²(x) - 5sin²(x)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 5cos²(x) - 5sin²(x)</p>
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<p>Thus, the derivative of the specified function is 5cos²(x) - 5sin²(x).</p>
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<p>Thus, the derivative of the specified function is 5cos²(x) - 5sin²(x).</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>AXB International School installed a pendulum. The displacement is represented by the function y = 5sin(x), where y represents the displacement of the pendulum at angle x. If x = π/6 radians, measure the rate of change of displacement.</p>
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<p>AXB International School installed a pendulum. The displacement is represented by the function y = 5sin(x), where y represents the displacement of the pendulum at angle x. If x = π/6 radians, measure the rate of change of displacement.</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 = 5sin(x) (rate of change of displacement)...(1)</p>
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<p>We have y = 5sin(x) (rate of change of displacement)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative 5sin(x): dy/dx = 5cos(x) Given x = π/6 (substitute this into the derivative) 5cos(π/6) = 5(√3/2)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative 5sin(x): dy/dx = 5cos(x) Given x = π/6 (substitute this into the derivative) 5cos(π/6) = 5(√3/2)</p>
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<p>Hence, we get the rate of change of displacement at x= π/6 as 5√3/2.</p>
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<p>Hence, we get the rate of change of displacement at x= π/6 as 5√3/2.</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 displacement at x= π/6 as 5√3/2, which means that at a given point, the displacement of the pendulum changes at this rate.</p>
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<p>We find the rate of change of displacement at x= π/6 as 5√3/2, which means that at a given point, the displacement of the pendulum changes at this rate.</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 = 5sin(x).</p>
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<p>Derive the second derivative of the function y = 5sin(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 = 5cos(x)...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = 5cos(x)...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [5cos(x)] = -5sin(x)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [5cos(x)] = -5sin(x)</p>
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<p>Therefore, the second derivative of the function y = 5sin(x) is -5sin(x).</p>
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<p>Therefore, the second derivative of the function y = 5sin(x) is -5sin(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. We differentiate the first derivative to find the second derivative.</p>
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<p>We use the step-by-step process, where we start with the first derivative. We differentiate the first derivative to find the second derivative.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>Prove: d/dx ((5sin(x))²) = 10sin(x)cos(x).</p>
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<p>Prove: d/dx ((5sin(x))²) = 10sin(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 = (5sin(x))² = [5sin(x)]² To differentiate, we use the chain rule: dy/dx = 2[5sin(x)]·d/dx [5sin(x)] = 2[5sin(x)]·5cos(x) = 10sin(x)cos(x) Hence proved.</p>
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<p>Let’s start using the chain rule: Consider y = (5sin(x))² = [5sin(x)]² To differentiate, we use the chain rule: dy/dx = 2[5sin(x)]·d/dx [5sin(x)] = 2[5sin(x)]·5cos(x) = 10sin(x)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 the function 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 the function 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 (5sin(x)/x)</p>
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<p>Solve: d/dx (5sin(x)/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 (5sin(x)/x) = (d/dx (5sin(x)).x - 5sin(x).d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (5sin(x)/x) = (d/dx (5sin(x)).x - 5sin(x).d/dx(x))/x²</p>
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<p>We will substitute d/dx (5sin(x)) = 5cos(x) and d/dx(x) = 1 = (5cos(x).x - 5sin(x).1)/x² = (5xcos(x) - 5sin(x))/x²</p>
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<p>We will substitute d/dx (5sin(x)) = 5cos(x) and d/dx(x) = 1 = (5cos(x).x - 5sin(x).1)/x² = (5xcos(x) - 5sin(x))/x²</p>
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<p>Therefore, d/dx (5sin(x)/x) = (5xcos(x) - 5sin(x))/x²</p>
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<p>Therefore, d/dx (5sin(x)/x) = (5xcos(x) - 5sin(x))/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 5sinx</h2>
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<h2>FAQs on the Derivative of 5sinx</h2>
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<h3>1.Find the derivative of 5sin(x).</h3>
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<h3>1.Find the derivative of 5sin(x).</h3>
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<p>Using the constant multiple rule on 5sin(x) gives 5cos(x).</p>
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<p>Using the constant multiple rule on 5sin(x) gives 5cos(x).</p>
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<h3>2.Can we use the derivative of 5sin(x) in real life?</h3>
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<h3>2.Can we use the derivative of 5sin(x) in real life?</h3>
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<p>Yes, we can use the derivative of 5sin(x) 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 5sin(x) 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 5sin(x) at the point where x = π/2?</h3>
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<h3>3.Is it possible to take the derivative of 5sin(x) at the point where x = π/2?</h3>
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<p>Yes, at x = π/2, the derivative 5cos(π/2) is 0 because cos(π/2) = 0.</p>
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<p>Yes, at x = π/2, the derivative 5cos(π/2) is 0 because cos(π/2) = 0.</p>
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<h3>4.What rule is used to differentiate 5sin(x)/x?</h3>
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<h3>4.What rule is used to differentiate 5sin(x)/x?</h3>
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<p>We use the quotient rule to differentiate 5sin(x)/x, d/dx (5sin(x)/x) = (5xcos(x) - 5sin(x))/x².</p>
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<p>We use the quotient rule to differentiate 5sin(x)/x, d/dx (5sin(x)/x) = (5xcos(x) - 5sin(x))/x².</p>
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<h3>5.Are the derivatives of 5sin(x) and sin⁻¹(x) the same?</h3>
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<h3>5.Are the derivatives of 5sin(x) and sin⁻¹(x) the same?</h3>
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<p>No, they are different. The derivative of 5sin(x) is 5cos(x), while the derivative of sin⁻¹(x) is 1/√(1-x²).</p>
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<p>No, they are different. The derivative of 5sin(x) is 5cos(x), while the derivative of sin⁻¹(x) is 1/√(1-x²).</p>
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<h2>Important Glossaries for the Derivative of 5sinx</h2>
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<h2>Important Glossaries for the Derivative of 5sinx</h2>
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<ul><li>Derivative: 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>Derivative: 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>Sine Function: A trigonometric function that is a common basis for waveforms.</li>
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</ul><ul><li>Sine Function: A trigonometric function that is a common basis for waveforms.</li>
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</ul><ul><li>Cosine Function: A trigonometric function derived as the derivative of sine and is used to describe oscillations. </li>
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</ul><ul><li>Cosine Function: A trigonometric function derived as the derivative of sine and is used to describe oscillations. </li>
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</ul><ul><li>Constant Multiple Rule: A rule that allows easy differentiation of functions multiplied by a constant.</li>
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</ul><ul><li>Constant Multiple Rule: A rule that allows easy differentiation of functions multiplied by a constant.</li>
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</ul><ul><li>First Derivative: It is the initial result of a function's differentiation, which gives us the rate of change of a specific function.</li>
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</ul><ul><li>First Derivative: It is the initial result of a function's differentiation, which gives us the rate of change of a specific function.</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>