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2026-01-01
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<p>Last updated on<strong>October 11, 2025</strong></p>
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<p>Last updated on<strong>October 11, 2025</strong></p>
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<p>We explore the derivative of arc functions such as arcsin, arccos, and arctan, which are crucial in understanding how the inverse trigonometric functions change concerning slight changes in x. These derivatives are widely used in various real-world applications, from calculating angles in physics to solving integrals in mathematics. We will now delve into the derivatives of inverse trig functions in detail.</p>
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<p>We explore the derivative of arc functions such as arcsin, arccos, and arctan, which are crucial in understanding how the inverse trigonometric functions change concerning slight changes in x. These derivatives are widely used in various real-world applications, from calculating angles in physics to solving integrals in mathematics. We will now delve into the derivatives of inverse trig functions in detail.</p>
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<h2>What is the Derivative of Arc Functions?</h2>
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<h2>What is the Derivative of Arc Functions?</h2>
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<p>The derivative<a>of</a>arc<a>functions</a>, namely arcsin, arccos, and arctan, are essential in<a>calculus</a>. They are represented as d/dx (arcsin x), d/dx (arccos x), and d/dx (arctan x), respectively. Each function has a well-defined derivative, indicating that they are differentiable within their domains.</p>
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<p>The derivative<a>of</a>arc<a>functions</a>, namely arcsin, arccos, and arctan, are essential in<a>calculus</a>. They are represented as d/dx (arcsin x), d/dx (arccos x), and d/dx (arctan x), respectively. Each function has a well-defined derivative, indicating that they are differentiable within their domains.</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>Arcsin Function: The inverse of the sine function.</p>
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<p>Arcsin Function: The inverse of the sine function.</p>
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<p>Arccos Function: The inverse of the cosine function.</p>
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<p>Arccos Function: The inverse of the cosine function.</p>
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<p>Arctan Function: The inverse of the tangent function.</p>
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<p>Arctan Function: The inverse of the tangent function.</p>
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<h2>Derivative of Arc Functions Formula</h2>
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<h2>Derivative of Arc Functions Formula</h2>
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<p>The derivatives of arc functions are represented as follows: d/dx (arcsin x) = 1/√(1-x²) d/dx (arccos x) = -1/√(1-x²) d/dx (arctan x) = 1/(1+x²)</p>
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<p>The derivatives of arc functions are represented as follows: d/dx (arcsin x) = 1/√(1-x²) d/dx (arccos x) = -1/√(1-x²) d/dx (arctan x) = 1/(1+x²)</p>
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<p>These<a>formulas</a>apply to their respective domains, where x is within the interval that ensures the<a>expressions</a>under the<a>square</a>root are non-negative.</p>
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<p>These<a>formulas</a>apply to their respective domains, where x is within the interval that ensures the<a>expressions</a>under the<a>square</a>root are non-negative.</p>
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<h2>Proofs of the Derivative of Arc Functions</h2>
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<h2>Proofs of the Derivative of Arc Functions</h2>
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<p>We can derive the derivatives of arc functions using proofs. To show this, we utilize trigonometric identities along with the rules of differentiation. Here are the methods used to prove the derivatives of arcsin, arccos, and arctan:</p>
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<p>We can derive the derivatives of arc functions using proofs. To show this, we utilize trigonometric identities along with the rules of differentiation. Here are the methods used to prove the derivatives of arcsin, arccos, and arctan:</p>
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<h2>Using Implicit Differentiation</h2>
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<h2>Using Implicit Differentiation</h2>
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<p>We will demonstrate that the differentiation of these arc functions results in their respective derivatives using implicit differentiation: Derivative of Arcsin x Let y = arcsin x, then sin y = x. Differentiating both sides with respect to x, we get: cos y · dy/dx = 1 dy/dx = 1/cos y Using the identity cos²y = 1 - sin²y and sin y = x, we have cos y = √(1 - x²). Thus, dy/dx = 1/√(1 - x²).</p>
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<p>We will demonstrate that the differentiation of these arc functions results in their respective derivatives using implicit differentiation: Derivative of Arcsin x Let y = arcsin x, then sin y = x. Differentiating both sides with respect to x, we get: cos y · dy/dx = 1 dy/dx = 1/cos y Using the identity cos²y = 1 - sin²y and sin y = x, we have cos y = √(1 - x²). Thus, dy/dx = 1/√(1 - x²).</p>
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<p>Derivative of Arccos x Let y = arccos x, then cos y = x. Differentiating both sides with respect to x, we get: -sin y · dy/dx = 1 dy/dx = -1/sin y Using the identity sin²y = 1 - cos²y and cos y = x, we have sin y = √(1 - x²). Thus, dy/dx = -1/√(1 - x²). Derivative of Arctan x Let y = arctan x, then tan y = x. Differentiating both sides with respect to x, we get: sec²y · dy/dx = 1 dy/dx = 1/sec²y Using the identity sec²y = 1 + tan²y and tan y = x, we have sec²y = 1 + x². Thus, dy/dx = 1/(1 + x²).</p>
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<p>Derivative of Arccos x Let y = arccos x, then cos y = x. Differentiating both sides with respect to x, we get: -sin y · dy/dx = 1 dy/dx = -1/sin y Using the identity sin²y = 1 - cos²y and cos y = x, we have sin y = √(1 - x²). Thus, dy/dx = -1/√(1 - x²). Derivative of Arctan x Let y = arctan x, then tan y = x. Differentiating both sides with respect to x, we get: sec²y · dy/dx = 1 dy/dx = 1/sec²y Using the identity sec²y = 1 + tan²y and tan y = x, we have sec²y = 1 + x². Thus, dy/dx = 1/(1 + x²).</p>
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<h2>Higher-Order Derivatives of Arc Functions</h2>
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<h2>Higher-Order Derivatives of Arc Functions</h2>
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<p>When a function is differentiated<a>multiple</a>times, the resulting derivatives are known as higher-order derivatives. Higher-order derivatives of arc functions provide insights akin to analyzing the acceleration (second derivative) and jerk (third derivative) in motion.</p>
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<p>When a function is differentiated<a>multiple</a>times, the resulting derivatives are known as higher-order derivatives. Higher-order derivatives of arc functions provide insights akin to analyzing the acceleration (second derivative) and jerk (third derivative) in motion.</p>
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<p>For the first derivative of a function, we write f′(x), indicating how the function changes or its slope at a certain point. The second derivative, f′′(x), results from differentiating the first derivative and continues in this pattern for higher-order derivatives.</p>
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<p>For the first derivative of a function, we write f′(x), indicating how the function changes or its slope at a certain point. The second derivative, f′′(x), results from differentiating the first derivative and continues in this pattern 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>Certain values of x may cause the derivative of arc functions to be undefined: For arcsin x and arccos x, the derivatives are undefined at x = ±1 because the<a>square root</a>in the<a>denominator</a>becomes zero.</p>
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<p>Certain values of x may cause the derivative of arc functions to be undefined: For arcsin x and arccos x, the derivatives are undefined at x = ±1 because the<a>square root</a>in the<a>denominator</a>becomes zero.</p>
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<p>For arctan x, there are no points where the derivative is undefined within its domain.</p>
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<p>For arctan x, there are no points where the derivative is undefined within its domain.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Arc Functions</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Arc Functions</h2>
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<p>Students often encounter errors when differentiating arc functions. Understanding the correct methods can resolve these mistakes. Here are common mistakes and how to avoid them:</p>
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<p>Students often encounter errors when differentiating arc functions. Understanding the correct methods can resolve these mistakes. Here are common mistakes and how to avoid 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 (arcsin x · x²)</p>
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<p>Calculate the derivative of (arcsin x · x²)</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) = arcsin x · x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = arcsin x and v = x². Let’s differentiate each term, u′ = d/dx (arcsin x) = 1/√(1-x²) v′ = d/dx (x²) = 2x Substituting into the given equation, f'(x) = (1/√(1-x²)) · x² + arcsin x · 2x Let’s simplify terms to get the final answer, f'(x) = x²/√(1-x²) + 2x arcsin x Thus, the derivative of the specified function is x²/√(1-x²) + 2x arcsin x.</p>
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<p>Here, we have f(x) = arcsin x · x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = arcsin x and v = x². Let’s differentiate each term, u′ = d/dx (arcsin x) = 1/√(1-x²) v′ = d/dx (x²) = 2x Substituting into the given equation, f'(x) = (1/√(1-x²)) · x² + arcsin x · 2x Let’s simplify terms to get the final answer, f'(x) = x²/√(1-x²) + 2x arcsin x Thus, the derivative of the specified function is x²/√(1-x²) + 2x arcsin 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 it into two parts.</p>
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<p>We find the derivative of the given function by dividing it into two parts.</p>
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<p>The first step involves finding the derivative of each part, then combining them using the product rule to get the final result.</p>
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<p>The first step involves finding the derivative of each part, 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 satellite's trajectory is modeled by the function y = arctan(x), where y represents the angle from the ground at a distance x. If x = 1, determine the rate of change of the angle.</p>
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<p>A satellite's trajectory is modeled by the function y = arctan(x), where y represents the angle from the ground at a distance x. If x = 1, determine the rate of change of the angle.</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 = arctan(x) (angle of the satellite)...(1) Now, we will differentiate equation (1) Take the derivative of arctan(x): dy/dx = 1/(1+x²) Given x = 1, substitute this into the derivative dy/dx = 1/(1+1²) dy/dx = 1/2 Hence, the rate of change of the angle at x = 1 is 1/2.</p>
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<p>We have y = arctan(x) (angle of the satellite)...(1) Now, we will differentiate equation (1) Take the derivative of arctan(x): dy/dx = 1/(1+x²) Given x = 1, substitute this into the derivative dy/dx = 1/(1+1²) dy/dx = 1/2 Hence, the rate of change of the angle at x = 1 is 1/2.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We calculate the rate of change of the angle at x = 1, indicating how the angle varies with respect to changes in distance.</p>
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<p>We calculate the rate of change of the angle at x = 1, indicating how the angle varies with respect to changes in distance.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function y = arcsin(x).</p>
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<p>Derive the second derivative of the function y = arcsin(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 = 1/√(1-x²)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/√(1-x²)] Using the chain rule, d²y/dx² = -(1-x²)^(-3/2) · (-2x) d²y/dx² = 2x/(1-x²)^(3/2) Therefore, the second derivative of the function y = arcsin(x) is 2x/(1-x²)^(3/2).</p>
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<p>The first step is to find the first derivative, dy/dx = 1/√(1-x²)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/√(1-x²)] Using the chain rule, d²y/dx² = -(1-x²)^(-3/2) · (-2x) d²y/dx² = 2x/(1-x²)^(3/2) Therefore, the second derivative of the function y = arcsin(x) is 2x/(1-x²)^(3/2).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We utilize the chain rule, where the first derivative is differentiated to find the second derivative.</p>
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<p>We utilize the chain rule, where the first derivative is differentiated to find the second derivative.</p>
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<p>We simplify using the chain rule and trigonometric identities to find the final answer.</p>
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<p>We simplify using the chain rule and trigonometric identities 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 (arctan²(x)) = 2 arctan(x)/(1+x²).</p>
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<p>Prove: d/dx (arctan²(x)) = 2 arctan(x)/(1+x²).</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 = arctan²(x) [arctan(x)]² To differentiate, we use the chain rule: dy/dx = 2 arctan(x) · d/dx [arctan(x)] Since the derivative of arctan(x) is 1/(1+x²), dy/dx = 2 arctan(x) · 1/(1+x²) Substituting y = arctan²(x), d/dx (arctan²(x)) = 2 arctan(x)/(1+x²) Hence proved.</p>
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<p>Let’s start using the chain rule: Consider y = arctan²(x) [arctan(x)]² To differentiate, we use the chain rule: dy/dx = 2 arctan(x) · d/dx [arctan(x)] Since the derivative of arctan(x) is 1/(1+x²), dy/dx = 2 arctan(x) · 1/(1+x²) Substituting y = arctan²(x), d/dx (arctan²(x)) = 2 arctan(x)/(1+x²) 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 use the chain rule to differentiate the equation.</p>
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<p>In this step-by-step process, we use the chain rule to differentiate the equation.</p>
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<p>Then, we replace arctan(x) with its derivative.</p>
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<p>Then, we replace arctan(x) with its derivative.</p>
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<p>Finally, we substitute y = arctan²(x) to derive the equation.</p>
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<p>Finally, we substitute y = arctan²(x) 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 (arcsin(x)/x)</p>
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<p>Solve: d/dx (arcsin(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 (arcsin(x)/x) = (d/dx (arcsin(x)) · x - arcsin(x) · d/dx(x))/x² Substitute d/dx (arcsin(x)) = 1/√(1-x²) and d/dx(x) = 1 (1/√(1-x²) · x - arcsin(x) · 1)/x² = (x/√(1-x²) - arcsin(x))/x² Therefore, d/dx (arcsin(x)/x) = (x/√(1-x²) - arcsin(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (arcsin(x)/x) = (d/dx (arcsin(x)) · x - arcsin(x) · d/dx(x))/x² Substitute d/dx (arcsin(x)) = 1/√(1-x²) and d/dx(x) = 1 (1/√(1-x²) · x - arcsin(x) · 1)/x² = (x/√(1-x²) - arcsin(x))/x² Therefore, d/dx (arcsin(x)/x) = (x/√(1-x²) - arcsin(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.</p>
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<p>In this process, we differentiate the given function using the quotient rule.</p>
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<p>As a final step, we simplify the equation to obtain the final result.</p>
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<p>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 Arc Functions</h2>
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<h2>FAQs on the Derivative of Arc Functions</h2>
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<h3>1.Find the derivative of arctan x.</h3>
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<h3>1.Find the derivative of arctan x.</h3>
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<p>The derivative of arctan x is 1/(1+x²), obtained using implicit differentiation.</p>
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<p>The derivative of arctan x is 1/(1+x²), obtained using implicit differentiation.</p>
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<h3>2.Can we use the derivative of arc functions in real life?</h3>
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<h3>2.Can we use the derivative of arc functions in real life?</h3>
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<p>Yes, derivatives of arc functions are applied in calculating angles and rates of change in fields such as physics, engineering, and computer graphics.</p>
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<p>Yes, derivatives of arc functions are applied in calculating angles and rates of change in fields such as physics, engineering, and computer graphics.</p>
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<h3>3.Is it possible to take the derivative of arcsin x at the point where x = 1?</h3>
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<h3>3.Is it possible to take the derivative of arcsin x at the point where x = 1?</h3>
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<p>No, x = 1 is a point where the derivative of arcsin x is undefined, as the expression under the square root becomes zero.</p>
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<p>No, x = 1 is a point where the derivative of arcsin x is undefined, as the expression under the square root becomes zero.</p>
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<h3>4.What rule is used to differentiate arcsin(x)/x?</h3>
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<h3>4.What rule is used to differentiate arcsin(x)/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate arcsin(x)/x, resulting in (x/√(1-x²) - arcsin(x))/x².</p>
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<p>We use the<a>quotient</a>rule to differentiate arcsin(x)/x, resulting in (x/√(1-x²) - arcsin(x))/x².</p>
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<h3>5.Are the derivatives of arcsin x and arccos x the same?</h3>
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<h3>5.Are the derivatives of arcsin x and arccos x the same?</h3>
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<p>No, they are different. The derivative of arcsin x is 1/√(1-x²), while the derivative of arccos x is -1/√(1-x²).</p>
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<p>No, they are different. The derivative of arcsin x is 1/√(1-x²), while the derivative of arccos x is -1/√(1-x²).</p>
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<h3>6.Can we find the derivative of the arctan x formula?</h3>
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<h3>6.Can we find the derivative of the arctan x formula?</h3>
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<p>To find, consider y = arctan x. Using implicit differentiation: d/dx (tan y) = d/dx(x) sec²y · dy/dx = 1 dy/dx = 1/sec²y = 1/(1+x²)</p>
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<p>To find, consider y = arctan x. Using implicit differentiation: d/dx (tan y) = d/dx(x) sec²y · dy/dx = 1 dy/dx = 1/sec²y = 1/(1+x²)</p>
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<h2>Important Glossaries for the Derivative of Arc Functions</h2>
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<h2>Important Glossaries for the Derivative of Arc Functions</h2>
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<ul><li><strong>Derivative:</strong>Indicates how the given function changes with respect to a slight change in x.</li>
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<ul><li><strong>Derivative:</strong>Indicates how the given function changes with respect to a slight change in x.</li>
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</ul><ul><li><strong>Inverse Trigonometric Functions:</strong>Functions that reverse the effect of sine, cosine, and tangent, known as arcsin, arccos, and arctan.</li>
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</ul><ul><li><strong>Inverse Trigonometric Functions:</strong>Functions that reverse the effect of sine, cosine, and tangent, known as arcsin, arccos, and arctan.</li>
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</ul><ul><li><strong>Implicit Differentiation:</strong>A technique used to find the derivative of a function defined implicitly.</li>
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</ul><ul><li><strong>Implicit Differentiation:</strong>A technique used to find the derivative of a function defined implicitly.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating the quotient of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating the quotient 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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<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>