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2026-01-01
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<p>Last updated on<strong>September 27, 2025</strong></p>
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<p>Last updated on<strong>September 27, 2025</strong></p>
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<p>We use the derivative of tan⁻¹(x), which is 1/(1 + x²), as a tool to understand how the inverse tangent function changes with respect to slight changes in x. Derivatives are essential for calculating rates of change in various real-life situations. We will now discuss the derivative of tan⁻¹(x) in detail.</p>
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<p>We use the derivative of tan⁻¹(x), which is 1/(1 + x²), as a tool to understand how the inverse tangent function changes with respect to slight changes in x. Derivatives are essential for calculating rates of change in various real-life situations. We will now discuss the derivative of tan⁻¹(x) in detail.</p>
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<h2>What is the Derivative of Tan⁻¹x?</h2>
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<h2>What is the Derivative of Tan⁻¹x?</h2>
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<p>We now understand the derivative of tan⁻¹x. It is commonly represented as d/dx (tan⁻¹x) or (tan⁻¹x)', and its value is 1/(1 + x²). The<a>function</a>tan⁻¹x has a clearly defined derivative, indicating it is differentiable for all<a>real numbers</a>.</p>
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<p>We now understand the derivative of tan⁻¹x. It is commonly represented as d/dx (tan⁻¹x) or (tan⁻¹x)', and its value is 1/(1 + x²). The<a>function</a>tan⁻¹x has a clearly defined derivative, indicating it is differentiable for all<a>real numbers</a>.</p>
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<p>The key concepts are mentioned below: Inverse Tangent Function: (tan⁻¹(x) is the inverse of tan(x)). Chain Rule: A rule for differentiating composite functions. Secant Function: sec(x) = 1/cos(x).</p>
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<p>The key concepts are mentioned below: Inverse Tangent Function: (tan⁻¹(x) is the inverse of tan(x)). Chain Rule: A rule for differentiating composite functions. Secant Function: sec(x) = 1/cos(x).</p>
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<h2>Derivative of Tan⁻¹x Formula</h2>
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<h2>Derivative of Tan⁻¹x Formula</h2>
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<p>The derivative of tan⁻¹x can be denoted as d/dx (tan⁻¹x) or (tan⁻¹x)'. The<a>formula</a>we use to differentiate tan⁻¹x is: d/dx (tan⁻¹x) = 1/(1 + x²)</p>
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<p>The derivative of tan⁻¹x can be denoted as d/dx (tan⁻¹x) or (tan⁻¹x)'. The<a>formula</a>we use to differentiate tan⁻¹x is: d/dx (tan⁻¹x) = 1/(1 + x²)</p>
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<p>The formula applies to all real x.</p>
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<p>The formula applies to all real x.</p>
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<h2>Proofs of the Derivative of Tan⁻¹x</h2>
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<h2>Proofs of the Derivative of Tan⁻¹x</h2>
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<p>We can derive the derivative of tan⁻¹x using proofs. To show this, we will use implicit differentiation and trigonometric identities.</p>
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<p>We can derive the derivative of tan⁻¹x using proofs. To show this, we will use implicit differentiation and trigonometric identities.</p>
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<p>There are several methods to prove this, such as:</p>
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<p>There are several methods to prove this, such as:</p>
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<ul><li>Using Implicit Differentiation </li>
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<ul><li>Using Implicit Differentiation </li>
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<li>Using Trigonometric Identities</li>
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<li>Using Trigonometric Identities</li>
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</ul><p>We will now demonstrate that the differentiation of tan⁻¹x results in 1/(1 + x²) using these methods:</p>
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</ul><p>We will now demonstrate that the differentiation of tan⁻¹x results in 1/(1 + x²) using these methods:</p>
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<h2><strong>Using Implicit Differentiation</strong></h2>
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<h2><strong>Using Implicit Differentiation</strong></h2>
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<p>Consider y = tan⁻¹x. This implies x = tan(y). Differentiating both sides with respect to x gives: 1 = sec²(y) · dy/dx Recall that sec²(y) = 1 + tan²(y). Thus, 1 = (1 + x²) dy/dx (since tan(y) = x) Therefore, dy/dx = 1/(1 + x²).</p>
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<p>Consider y = tan⁻¹x. This implies x = tan(y). Differentiating both sides with respect to x gives: 1 = sec²(y) · dy/dx Recall that sec²(y) = 1 + tan²(y). Thus, 1 = (1 + x²) dy/dx (since tan(y) = x) Therefore, dy/dx = 1/(1 + x²).</p>
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<h2><strong>Using Trigonometric Identities</strong></h2>
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<h2><strong>Using Trigonometric Identities</strong></h2>
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<p>We know the identity tan²(y) + 1 = sec²(y). Therefore, using implicit differentiation, we substitute tan(y) = x, resulting in: dy/dx = 1/(1 + x²).</p>
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<p>We know the identity tan²(y) + 1 = sec²(y). Therefore, using implicit differentiation, we substitute tan(y) = x, resulting in: dy/dx = 1/(1 + x²).</p>
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<h2>Higher-Order Derivatives of Tan⁻¹x</h2>
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<h2>Higher-Order Derivatives of Tan⁻¹x</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 bit complex. To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives help analyze functions like tan⁻¹(x).</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 bit complex. To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives help analyze functions like tan⁻¹(x).</p>
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<p>For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>For the nth Derivative of tan⁻¹(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 (continuing for higher-order derivatives).</p>
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<p>For the nth Derivative of tan⁻¹(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 (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 = 0, the derivative of tan⁻¹x = 1/(1 + 0²), which is 1.</p>
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<p>When x = 0, the derivative of tan⁻¹x = 1/(1 + 0²), which is 1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Tan⁻¹x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Tan⁻¹x</h2>
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<p>Students frequently make mistakes when differentiating tan⁻¹x. 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 tan⁻¹x. 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 tan⁻¹(3x).</p>
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<p>Calculate the derivative of tan⁻¹(3x).</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 y = tan⁻¹(3x). Using the chain rule, dy/dx = d/dx [tan⁻¹(3x)] = 3/(1 + (3x)²) = 3/(1 + 9x²).</p>
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<p>Let y = tan⁻¹(3x). Using the chain rule, dy/dx = d/dx [tan⁻¹(3x)] = 3/(1 + (3x)²) = 3/(1 + 9x²).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the derivative by recognizing tan⁻¹(3x) as a composite function.</p>
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<p>We find the derivative by recognizing tan⁻¹(3x) as a composite function.</p>
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<p>We apply the chain rule to differentiate the outer function tan⁻¹ and then the inner function 3x, combining the results.</p>
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<p>We apply the chain rule to differentiate the outer function tan⁻¹ and then the inner function 3x, combining the results.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>The height of a water tank is given by the function h(x) = tan⁻¹(x), where x is the time in hours. Find the rate of change of the height when x = 1 hour.</p>
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<p>The height of a water tank is given by the function h(x) = tan⁻¹(x), where x is the time in hours. Find the rate of change of the height when x = 1 hour.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Given h(x) = tan⁻¹(x), the rate of change is the derivative: dh/dx = 1/(1 + x²).</p>
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<p>Given h(x) = tan⁻¹(x), the rate of change is the derivative: dh/dx = 1/(1 + x²).</p>
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<p>Substitute x = 1: dh/dx = 1/(1 + 1²) = 1/2.</p>
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<p>Substitute x = 1: dh/dx = 1/(1 + 1²) = 1/2.</p>
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<p>Thus, the rate of change of the height at x = 1 hour is 1/2 units per hour.</p>
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<p>Thus, the rate of change of the height at x = 1 hour is 1/2 units per hour.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>To find the rate of change, we differentiate the function h(x) = tan⁻¹(x) and evaluate it at x = 1. The result indicates how the height changes with respect to time at that specific moment.</p>
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<p>To find the rate of change, we differentiate the function h(x) = tan⁻¹(x) and evaluate it at x = 1. The result indicates how the height changes with respect to time 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 = tan⁻¹(x).</p>
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<p>Derive the second derivative of the function y = tan⁻¹(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>First, find the first derivative: dy/dx = 1/(1 + x²).Now differentiate again to find the second derivative: d²y/dx² = d/dx [1/(1 + x²)] = -2x/(1 + x²)². Therefore, the second derivative of the function y = tan⁻¹(x) is -2x/(1 + x²)².</p>
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<p>First, find the first derivative: dy/dx = 1/(1 + x²).Now differentiate again to find the second derivative: d²y/dx² = d/dx [1/(1 + x²)] = -2x/(1 + x²)². Therefore, the second derivative of the function y = tan⁻¹(x) is -2x/(1 + x²)².</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use implicit differentiation to find the first derivative, then differentiate once more, applying the quotient rule to obtain the second derivative.</p>
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<p>We use implicit differentiation to find the first derivative, then differentiate once more, applying the quotient rule to obtain the second derivative.</p>
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<p>Simplifying gives the final answer.</p>
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<p>Simplifying gives 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 (tan⁻¹(x²)) = 2x/(1 + x⁴).</p>
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<p>Prove: d/dx (tan⁻¹(x²)) = 2x/(1 + x⁴).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Consider y = tan⁻¹(x²). Using the chain rule: dy/dx = d/dx [tan⁻¹(x²)] = 1/(1 + (x²)²) · d/dx (x²) = 1/(1 + x⁴) · 2x = 2x/(1 + x⁴). Hence proved.</p>
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<p>Consider y = tan⁻¹(x²). Using the chain rule: dy/dx = d/dx [tan⁻¹(x²)] = 1/(1 + (x²)²) · d/dx (x²) = 1/(1 + x⁴) · 2x = 2x/(1 + x⁴). Hence proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we applied the chain rule to differentiate tan⁻¹(x²).</p>
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<p>In this process, we applied the chain rule to differentiate tan⁻¹(x²).</p>
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<p>We first differentiate the outer function tan⁻¹ and then multiply by the derivative of the inner function x², resulting in the final expression.</p>
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<p>We first differentiate the outer function tan⁻¹ and then multiply by the derivative of the inner function x², resulting in the final expression.</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 (tan⁻¹(x²)/x).</p>
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<p>Solve: d/dx (tan⁻¹(x²)/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 (tan⁻¹(x²)/x) = (x · d/dx (tan⁻¹(x²)) - tan⁻¹(x²) · d/dx (x))/x². Substitute d/dx (tan⁻¹(x²)) = 2x/(1 + x⁴): = (x · (2x/(1 + x⁴)) - tan⁻¹(x²) · 1)/x² = (2x²/(1 + x⁴) - tan⁻¹(x²))/x². Therefore, d/dx (tan⁻¹(x²)/x) = (2x²/(1 + x⁴) - tan⁻¹(x²))/x².</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (tan⁻¹(x²)/x) = (x · d/dx (tan⁻¹(x²)) - tan⁻¹(x²) · d/dx (x))/x². Substitute d/dx (tan⁻¹(x²)) = 2x/(1 + x⁴): = (x · (2x/(1 + x⁴)) - tan⁻¹(x²) · 1)/x² = (2x²/(1 + x⁴) - tan⁻¹(x²))/x². Therefore, d/dx (tan⁻¹(x²)/x) = (2x²/(1 + x⁴) - tan⁻¹(x²))/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>We apply the derivative of tan⁻¹(x²) and simplify the equation to obtain the final result.</p>
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<p>We apply the derivative of tan⁻¹(x²) and 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 Tan⁻¹x</h2>
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<h2>FAQs on the Derivative of Tan⁻¹x</h2>
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<h3>1.Find the derivative of tan⁻¹x.</h3>
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<h3>1.Find the derivative of tan⁻¹x.</h3>
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<p>Using implicit differentiation, we find that: d/dx (tan⁻¹x) = 1/(1 + x²).</p>
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<p>Using implicit differentiation, we find that: d/dx (tan⁻¹x) = 1/(1 + x²).</p>
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<h3>2.Can we use the derivative of tan⁻¹x in real life?</h3>
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<h3>2.Can we use the derivative of tan⁻¹x in real life?</h3>
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<p>Yes, the derivative of tan⁻¹x is used in real-life scenarios involving angles and rates of change, particularly in fields such as physics and engineering.</p>
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<p>Yes, the derivative of tan⁻¹x is used in real-life scenarios involving angles and rates of change, particularly in fields such as physics and engineering.</p>
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<h3>3.What happens to the derivative of tan⁻¹x as x becomes very large?</h3>
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<h3>3.What happens to the derivative of tan⁻¹x as x becomes very large?</h3>
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<p>As x becomes very large, the derivative of tan⁻¹x, 1/(1 + x²), approaches zero. This indicates that the function becomes less sensitive to changes in x as x increases.</p>
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<p>As x becomes very large, the derivative of tan⁻¹x, 1/(1 + x²), approaches zero. This indicates that the function becomes less sensitive to changes in x as x increases.</p>
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<h3>4.What rule is used to differentiate tan⁻¹(x²)?</h3>
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<h3>4.What rule is used to differentiate tan⁻¹(x²)?</h3>
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<p>We use the chain rule to differentiate tan⁻¹(x²), d/dx (tan⁻¹(x²)) = 2x/(1 + x⁴).</p>
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<p>We use the chain rule to differentiate tan⁻¹(x²), d/dx (tan⁻¹(x²)) = 2x/(1 + x⁴).</p>
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<h3>5.Are the derivatives of tan x and tan⁻¹x the same?</h3>
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<h3>5.Are the derivatives of tan x and tan⁻¹x the same?</h3>
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<p>No, they are different. The derivative of tan x is equal to sec²x, while the derivative of tan⁻¹x is 1/(1 + x²).</p>
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<p>No, they are different. The derivative of tan x is equal to sec²x, while the derivative of tan⁻¹x is 1/(1 + x²).</p>
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<h2>Important Glossaries for the Derivative of Tan⁻¹x</h2>
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<h2>Important Glossaries for the Derivative of Tan⁻¹x</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the 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 function changes in response to a slight change in x.</li>
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</ul><ul><li><strong>Inverse Tangent Function:</strong>The function tan⁻¹(x) is the inverse of the tangent function and is used to find the angle whose tangent is x.</li>
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</ul><ul><li><strong>Inverse Tangent Function:</strong>The function tan⁻¹(x) is the inverse of the tangent function and is used to find the angle whose tangent is x.</li>
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</ul><ul><li><strong>Implicit Differentiation:</strong>A technique used to differentiate equations where the dependent variable is not isolated on one side of the equation.</li>
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</ul><ul><li><strong>Implicit Differentiation:</strong>A technique used to differentiate equations where the dependent variable is not isolated on one side of the equation.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate composite functions, where one function is nested inside another.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate composite functions, where one function is nested inside another.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate functions of the form u/v, where both u and v are differentiable functions of x.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate functions of the form u/v, where both u and v are differentiable functions of x.</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>