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2026-01-01
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>We use the derivatives of trigonometric functions squared to understand how these functions change in response to a slight change in x. Derivatives help us in various real-life applications, such as calculating rates of change. We will now discuss the derivatives of trigonometric functions squared in detail.</p>
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<p>We use the derivatives of trigonometric functions squared to understand how these functions change in response to a slight change in x. Derivatives help us in various real-life applications, such as calculating rates of change. We will now discuss the derivatives of trigonometric functions squared in detail.</p>
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<h2>What is the Derivative of a Trig Function Squared?</h2>
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<h2>What is the Derivative of a Trig Function Squared?</h2>
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<p>We now explore the derivative<a>of</a>trig<a>functions</a>squared. For example, the derivative of (tan x)² is commonly represented as d/dx ((tan x)²) or ((tan x)²)', and its value is 2 tan x sec²x.</p>
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<p>We now explore the derivative<a>of</a>trig<a>functions</a>squared. For example, the derivative of (tan x)² is commonly represented as d/dx ((tan x)²) or ((tan x)²)', and its value is 2 tan x sec²x.</p>
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<p>Trigonometric functions like tan x have well-defined derivatives within their domains, indicating they are differentiable. The key concepts are mentioned below:</p>
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<p>Trigonometric functions like tan x have well-defined derivatives within their domains, indicating they are differentiable. The key concepts are mentioned below:</p>
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<p><strong>Tangent Function:</strong>(tan(x) = sin(x)/cos(x)).</p>
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<p><strong>Tangent Function:</strong>(tan(x) = sin(x)/cos(x)).</p>
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<p><strong>Square of a Function:</strong>When a function is squared, its derivative involves the original function and its derivative.</p>
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<p><strong>Square of a Function:</strong>When a function is squared, its derivative involves the original function and its derivative.</p>
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<p><strong>Secant Function:</strong>sec(x) = 1/cos(x).</p>
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<p><strong>Secant Function:</strong>sec(x) = 1/cos(x).</p>
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<h2>Derivative of Trig Functions Squared Formula</h2>
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<h2>Derivative of Trig Functions Squared Formula</h2>
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<p>The derivative of a squared trig function like (tan x)² can be denoted as d/dx ((tan x)²) or ((tan x)²)'.</p>
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<p>The derivative of a squared trig function like (tan x)² can be denoted as d/dx ((tan x)²) or ((tan x)²)'.</p>
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<p>The<a>formula</a>we use is: d/dx ((tan x)²) = 2 tan x sec²x</p>
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<p>The<a>formula</a>we use is: d/dx ((tan x)²) = 2 tan x sec²x</p>
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<p>This formula applies to all x where the original function is defined and differentiable.</p>
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<p>This formula applies to all x where the original function is defined and differentiable.</p>
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<h2>Proofs of the Derivative of Trig Functions Squared</h2>
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<h2>Proofs of the Derivative of Trig Functions Squared</h2>
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<p>We can derive the derivative of a squared trig function using proofs and rules of differentiation. There are several methods to prove this, such as:</p>
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<p>We can derive the derivative of a squared trig function using proofs and rules of differentiation. There are several methods to prove this, such as:</p>
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<ol><li>Using the Chain Rule</li>
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<ol><li>Using the Chain Rule</li>
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<li>Using the Product Rule</li>
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<li>Using the Product Rule</li>
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</ol><p>We will now demonstrate the differentiation of (tan x)² using these methods:</p>
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</ol><p>We will now demonstrate the differentiation of (tan x)² using these methods:</p>
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<h3>Using the Chain Rule</h3>
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<h3>Using the Chain Rule</h3>
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<p>To prove the differentiation of (tan x)² using the chain rule, consider y = (tan x)².</p>
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<p>To prove the differentiation of (tan x)² using the chain rule, consider y = (tan x)².</p>
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<p>We use the formula: dy/dx = 2 tan x · d/dx (tan x)</p>
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<p>We use the formula: dy/dx = 2 tan x · d/dx (tan x)</p>
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<p>Since the derivative of tan x is sec²x, dy/dx = 2 tan x · sec²x</p>
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<p>Since the derivative of tan x is sec²x, dy/dx = 2 tan x · sec²x</p>
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<p>Hence, d/dx ((tan x)²) = 2 tan x sec²x.</p>
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<p>Hence, d/dx ((tan x)²) = 2 tan x sec²x.</p>
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<h3>Using the Product Rule</h3>
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<h3>Using the Product Rule</h3>
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<p>We will now prove the derivative of (tan x)² using the<a>product</a>rule.</p>
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<p>We will now prove the derivative of (tan x)² using the<a>product</a>rule.</p>
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<p>Here, we consider: (tan x)² = (tan x) · (tan x)</p>
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<p>Here, we consider: (tan x)² = (tan x) · (tan x)</p>
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<p>Using the product rule formula: d/dx [u · v] = u' · v + u · v'</p>
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<p>Using the product rule formula: d/dx [u · v] = u' · v + u · v'</p>
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<p>Let u = tan x and v = tan x, u' = sec²x and v' = sec²x</p>
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<p>Let u = tan x and v = tan x, u' = sec²x and v' = sec²x</p>
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<p>d/dx ((tan x)²) = (sec²x) · (tan x) + (tan x) · (sec²x) = 2 tan x sec²x</p>
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<p>d/dx ((tan x)²) = (sec²x) · (tan x) + (tan x) · (sec²x) = 2 tan x sec²x</p>
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<p>Thus, the derivative of (tan x)² is 2 tan x sec²x.</p>
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<p>Thus, the derivative of (tan x)² is 2 tan x sec²x.</p>
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<h2>Higher-Order Derivatives of Trig Functions Squared</h2>
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<h2>Higher-Order Derivatives of Trig Functions Squared</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 complex.</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 complex.</p>
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<p>For instance, the first derivative indicates how the function changes, while the second derivative measures the<a>rate</a>of that change. Higher-order derivatives help in understanding functions like (tan x)².</p>
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<p>For instance, the first derivative indicates how the function changes, while the second derivative measures the<a>rate</a>of that change. Higher-order derivatives help in understanding functions like (tan x)².</p>
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<p>For the first derivative, we write f′(x), which indicates how the function changes at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x). Similarly, the third derivative, f′′′(x), is the derivative of the second derivative.</p>
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<p>For the first derivative, we write f′(x), which indicates how the function changes at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x). Similarly, the third derivative, f′′′(x), is the derivative of the second derivative.</p>
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<p>For the nth Derivative of (tan x)², we use fⁿ(x) for the nth derivative, indicating the change in the rate of change.</p>
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<p>For the nth Derivative of (tan x)², we use fⁿ(x) for the nth derivative, indicating 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 undefined because tan(x) has a vertical asymptote there. When x is 0, the derivative of (tan x)² = 2 tan(0) sec²(0), which is 0.</p>
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<p>When x is π/2, the derivative is undefined because tan(x) has a vertical asymptote there. When x is 0, the derivative of (tan x)² = 2 tan(0) sec²(0), which is 0.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Trig Functions Squared</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Trig Functions Squared</h2>
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<p>Students frequently make mistakes when differentiating trig functions squared. 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 trig functions squared. 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 x)² · sec²x)</p>
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<p>Calculate the derivative of ((sin x)² · sec²x)</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 x)² · sec²x.</p>
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<p>Here, we have f(x) = (sin x)² · sec²x.</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = (sin x)² and v = sec²x.</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = (sin x)² and v = sec²x.</p>
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<p>Let’s differentiate each term, u′ = d/dx ((sin x)²) = 2 sin x cos x v′ = d/dx (sec²x) = 2 sec²x tan x</p>
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<p>Let’s differentiate each term, u′ = d/dx ((sin x)²) = 2 sin x cos x v′ = d/dx (sec²x) = 2 sec²x tan x</p>
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<p>Substituting into the given equation, f'(x) = (2 sin x cos x) · (sec²x) + ((sin x)²) · (2 sec²x tan x)</p>
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<p>Substituting into the given equation, f'(x) = (2 sin x cos x) · (sec²x) + ((sin x)²) · (2 sec²x tan x)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 2 sin x cos x sec²x + 2 (sin x)² sec²x tan x</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 2 sin x cos x sec²x + 2 (sin x)² sec²x tan x</p>
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<p>Thus, the derivative of the specified function is 2 sin x cos x sec²x + 2 (sin x)² sec²x tan x.</p>
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<p>Thus, the derivative of the specified function is 2 sin x cos x sec²x + 2 (sin x)² sec²x tan 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>A company measures the brightness of a light source using the function y = (cos(x))², where y represents the intensity at angle x. If x = π/6 radians, calculate the rate of change in intensity.</p>
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<p>A company measures the brightness of a light source using the function y = (cos(x))², where y represents the intensity at angle x. If x = π/6 radians, calculate the rate of change in intensity.</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 = (cos(x))² (intensity of the light)...(1)</p>
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<p>We have y = (cos(x))² (intensity of the light)...(1)</p>
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<p>Now, we will differentiate equation (1) Take the derivative of (cos(x))²: dy/dx = 2 cos(x)(-sin(x))</p>
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<p>Now, we will differentiate equation (1) Take the derivative of (cos(x))²: dy/dx = 2 cos(x)(-sin(x))</p>
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<p>Substitute x = π/6 into the derivative: dy/dx = 2 cos(π/6)(-sin(π/6)) dy/dx = 2(√3/2)(-1/2) dy/dx = -√3/2</p>
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<p>Substitute x = π/6 into the derivative: dy/dx = 2 cos(π/6)(-sin(π/6)) dy/dx = 2(√3/2)(-1/2) dy/dx = -√3/2</p>
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<p>Hence, the rate of change in intensity at x = π/6 is -√3/2.</p>
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<p>Hence, the rate of change in intensity at x = π/6 is -√3/2.</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 light intensity at x = π/6 by differentiating the squared function and substituting the given angle to find the result.</p>
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<p>We calculate the rate of change of light intensity at x = π/6 by differentiating the squared function and substituting the given angle to find the result.</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(x))².</p>
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<p>Derive the second derivative of the function y = (sin(x))².</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>The first step is to find the first derivative: dy/dx = 2 sin(x) cos(x)...(1)</p>
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<p>The first step is to find the first derivative: dy/dx = 2 sin(x) cos(x)...(1)</p>
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<p>Now, differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2 sin(x) cos(x)]</p>
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<p>Now, differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2 sin(x) cos(x)]</p>
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<p>Using the product rule, d²y/dx² = 2[cos(x) cos(x) - sin(x) sin(x)] = 2[cos²(x) - sin²(x)]</p>
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<p>Using the product rule, d²y/dx² = 2[cos(x) cos(x) - sin(x) sin(x)] = 2[cos²(x) - sin²(x)]</p>
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<p>Therefore, the second derivative of the function y = (sin(x))² is 2[cos²(x) - sin²(x)].</p>
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<p>Therefore, the second derivative of the function y = (sin(x))² is 2[cos²(x) - sin²(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 to find the first derivative and then apply the product rule to differentiate further, simplifying to get the second derivative.</p>
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<p>We use the step-by-step process to find the first derivative and then apply the product rule to differentiate further, simplifying to get 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 ((cos(x))²) = -2 cos(x) sin(x).</p>
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<p>Prove: d/dx ((cos(x))²) = -2 cos(x) sin(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 = (cos(x))² = [cos(x)]²</p>
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<p>Let’s start using the chain rule: Consider y = (cos(x))² = [cos(x)]²</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2 cos(x) · d/dx [cos(x)]</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2 cos(x) · d/dx [cos(x)]</p>
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<p>Since the derivative of cos(x) is -sin(x), dy/dx = 2 cos(x) · (-sin(x)) dy/dx = -2 cos(x) sin(x) Hence proved.</p>
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<p>Since the derivative of cos(x) is -sin(x), dy/dx = 2 cos(x) · (-sin(x)) dy/dx = -2 cos(x) sin(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 replaced cos(x) with its derivative 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 replaced cos(x) with its derivative 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 ((tan x)²/x)</p>
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<p>Solve: d/dx ((tan x)²/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:</p>
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<p>To differentiate the function, we use the quotient rule:</p>
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<p>d/dx ((tan x)²/x) = (d/dx ((tan x)²) · x - (tan x)² · d/dx(x))/x²</p>
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<p>d/dx ((tan x)²/x) = (d/dx ((tan x)²) · x - (tan x)² · d/dx(x))/x²</p>
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<p>Substitute d/dx ((tan x)²) = 2 tan x sec²x and d/dx(x) = 1 = (2 tan x sec²x · x - (tan x)² · 1) / x² = (2 x tan x sec²x - (tan x)²) / x²</p>
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<p>Substitute d/dx ((tan x)²) = 2 tan x sec²x and d/dx(x) = 1 = (2 tan x sec²x · x - (tan x)² · 1) / x² = (2 x tan x sec²x - (tan x)²) / x²</p>
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<p>Therefore, d/dx ((tan x)²/x) = (2 x tan x sec²x - tan²x) / x²</p>
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<p>Therefore, d/dx ((tan x)²/x) = (2 x tan x sec²x - tan²x) / 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 Trig Functions Squared</h2>
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<h2>FAQs on the Derivative of Trig Functions Squared</h2>
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<h3>1.How do you find the derivative of (tan x)²?</h3>
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<h3>1.How do you find the derivative of (tan x)²?</h3>
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<p>Using the chain rule on (tan x)² gives us: d/dx ((tan x)²) = 2 tan x sec²x</p>
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<p>Using the chain rule on (tan x)² gives us: d/dx ((tan x)²) = 2 tan x sec²x</p>
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<h3>2.Can the derivative of a trig function squared be applied in real life?</h3>
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<h3>2.Can the derivative of a trig function squared be applied in real life?</h3>
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<p>Yes, derivatives of trig functions squared can be applied in real life to calculate rates of change in various fields, such as physics and engineering.</p>
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<p>Yes, derivatives of trig functions squared can be applied in real life to calculate rates of change in various fields, such as physics and engineering.</p>
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<h3>3.Can you differentiate (tan x)² at x = π/2?</h3>
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<h3>3.Can you differentiate (tan x)² at x = π/2?</h3>
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<p>No, π/2 is a point where tan x is undefined, so it is impossible to take the derivative at these points (since the function does not exist there).</p>
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<p>No, π/2 is a point where tan x is undefined, so it is impossible to take the derivative at these points (since the function does not exist there).</p>
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<h3>4.What rule is used to differentiate (tan x)²/x?</h3>
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<h3>4.What rule is used to differentiate (tan x)²/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate (tan x)²/x: d/dx ((tan x)²/x) = (2 x tan x sec²x - tan²x) / x².</p>
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<p>We use the<a>quotient</a>rule to differentiate (tan x)²/x: d/dx ((tan x)²/x) = (2 x tan x sec²x - tan²x) / x².</p>
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<h3>5.Are the derivatives of (tan x)² and tan⁻¹x the same?</h3>
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<h3>5.Are the derivatives of (tan x)² and tan⁻¹x the same?</h3>
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<p>No, they are different. The derivative of (tan x)² is 2 tan x 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)² is 2 tan x sec²x, while the derivative of tan⁻¹x is 1/(1 + x²).</p>
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<h2>Important Glossaries for the Derivative of Trig Functions Squared</h2>
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<h2>Important Glossaries for the Derivative of Trig Functions Squared</h2>
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<ul><li><strong>Derivative:</strong>A 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>A 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>Trig Function:</strong>A function based on angles, such as sine, cosine, or tangent, and their squares.</li>
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</ul><ul><li><strong>Trig Function:</strong>A function based on angles, such as sine, cosine, or tangent, and their squares.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for finding the derivative of a composite function.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for finding the derivative of a composite function.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate products of two or more functions.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used to differentiate products of two or more functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate a function that is the ratio of two differentiable functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate 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>