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2026-01-01
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2026-02-28
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<p>175 Learners</p>
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<p>211 Learners</p>
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<p>Last updated on<strong>September 9, 2025</strong></p>
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<p>Last updated on<strong>January 16, 2026</strong></p>
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<p>We use the derivative of y², which is 2y(dy/dx), as a tool to understand how the function changes in response to a small change in x. Derivatives are crucial in calculating rates of change in various contexts. We will now discuss the derivative of y² in detail.</p>
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<p>We use the derivative of y², which is 2y(dy/dx), as a tool to understand how the function changes in response to a small change in x. Derivatives are crucial in calculating rates of change in various contexts. We will now discuss the derivative of y² in detail.</p>
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<h2>What is the Derivative of y²?</h2>
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<h2>What is the Derivative of y²?</h2>
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<p>We now understand the derivative<a>of</a>y².</p>
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<p>We now understand the derivative<a>of</a>y².</p>
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<p>It is commonly represented as d/dx (y²) or (y²)', and its value is 2y(dy/dx).</p>
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<p>It is commonly represented as d/dx (y²) or (y²)', and its value is 2y(dy/dx).</p>
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<p>The<a>function</a>y² has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>The<a>function</a>y² has a clearly defined derivative, indicating it is differentiable within its domain.</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>Square Function: (y²) represents the<a>square</a>of y.</p>
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<p>Square Function: (y²) represents the<a>square</a>of y.</p>
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<p>Chain Rule: Rule for differentiating y² (since it involves y, which is a function of x).</p>
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<p>Chain Rule: Rule for differentiating y² (since it involves y, which is a function of x).</p>
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<p>Differentiation: Process of finding the derivative of a function.</p>
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<p>Differentiation: Process of finding the derivative of a function.</p>
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<h2>Derivative of y² Formula</h2>
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<h2>Derivative of y² Formula</h2>
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<p>The derivative of y² can be denoted as d/dx (y²) or (y²)'. The<a>formula</a>we use to differentiate y² is: d/dx (y²) = 2y(dy/dx) The formula applies to all x where y is differentiable.</p>
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<p>The derivative of y² can be denoted as d/dx (y²) or (y²)'. The<a>formula</a>we use to differentiate y² is: d/dx (y²) = 2y(dy/dx) The formula applies to all x where y is differentiable.</p>
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<h2>Proofs of the Derivative of y²</h2>
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<h2>Proofs of the Derivative of y²</h2>
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<p>We can derive the derivative of y² using proofs.</p>
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<p>We can derive the derivative of y² using proofs.</p>
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<p>To show this, we will use the chain rule of differentiation.</p>
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<p>To show this, we will use the chain rule of differentiation.</p>
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<p>The method we use is as follows:</p>
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<p>The method we use is as follows:</p>
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<p>Using Chain Rule</p>
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<p>Using Chain Rule</p>
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<p>To prove the differentiation of y² using the chain rule, we use the formula:</p>
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<p>To prove the differentiation of y² using the chain rule, we use the formula:</p>
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<p>Let y = f(x) be a function of x.</p>
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<p>Let y = f(x) be a function of x.</p>
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<p>Then y² = [f(x)]². Using the chain rule: d/dx [y²] = 2y(dy/dx)</p>
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<p>Then y² = [f(x)]². Using the chain rule: d/dx [y²] = 2y(dy/dx)</p>
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<p>We differentiate the outer function and multiply by the derivative of the inner function.</p>
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<p>We differentiate the outer function and multiply by the derivative of the inner function.</p>
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<p>Let's demonstrate that the differentiation of y² results in 2y(dy/dx) using this method:</p>
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<p>Let's demonstrate that the differentiation of y² results in 2y(dy/dx) using this method:</p>
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<p>Consider y² = (y)², where y = f(x). d/dx (y²) = 2y(dy/dx)</p>
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<p>Consider y² = (y)², where y = f(x). d/dx (y²) = 2y(dy/dx)</p>
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<p>The derivative of y² with respect to x is derived by using the chain rule, multiplying the derivative of the outer function 2y by the derivative of the inner function (dy/dx), which results in 2y(dy/dx).</p>
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<p>The derivative of y² with respect to x is derived by using the chain rule, multiplying the derivative of the outer function 2y by the derivative of the inner function (dy/dx), which results in 2y(dy/dx).</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h2>Higher-Order Derivatives of y²</h2>
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<h2>Higher-Order Derivatives of y²</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, consider an object 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, consider an object 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 y².</p>
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<p>Higher-order derivatives make it easier to understand functions like y².</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.</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.</p>
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<p>The second derivative is derived from the first derivative, which is denoted using f′′(x).</p>
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<p>The second derivative is derived from the first derivative, which is denoted using f′′(x).</p>
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<p>Similarly, the third derivative, f′′′(x), is the result of the second derivative and this pattern continues.</p>
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<p>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 y², 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 y², 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 y = 0, the derivative is 0 because any<a>constant</a><a>multiple</a>of 0 is 0. When y is a constant, the derivative of y² = 0, since the derivative of a constant is 0.</p>
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<p>When y = 0, the derivative is 0 because any<a>constant</a><a>multiple</a>of 0 is 0. When y is a constant, the derivative of y² = 0, since the derivative of a constant is 0.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of y²</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of y²</h2>
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<p>Students frequently make mistakes when differentiating y². 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 y². 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 (y² · sin y).</p>
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<p>Calculate the derivative of (y² · sin y).</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) = y² · sin y.</p>
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<p>Here, we have f(x) = y² · sin y.</p>
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<p>Using the product rule, f'(x) = u′v + uv′</p>
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<p>Using the product rule, f'(x) = u′v + uv′</p>
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<p>In the given equation, u = y² and v = sin y.</p>
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<p>In the given equation, u = y² and v = sin y.</p>
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<p>Let’s differentiate each term, u′ = d/dx (y²) = 2y(dy/dx) v′ = d/dx (sin y) = cos y(dy/dx)</p>
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<p>Let’s differentiate each term, u′ = d/dx (y²) = 2y(dy/dx) v′ = d/dx (sin y) = cos y(dy/dx)</p>
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<p>Substituting into the given equation, f'(x) = (2y(dy/dx)) · (sin y) + (y²) · (cos y(dy/dx))</p>
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<p>Substituting into the given equation, f'(x) = (2y(dy/dx)) · (sin y) + (y²) · (cos y(dy/dx))</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 2y sin y(dy/dx) + y² cos y(dy/dx)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 2y sin y(dy/dx) + y² cos y(dy/dx)</p>
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<p>Thus, the derivative of the specified function is 2y sin y(dy/dx) + y² cos y(dy/dx).</p>
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<p>Thus, the derivative of the specified function is 2y sin y(dy/dx) + y² cos y(dy/dx).</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 balloon rises such that its height is represented by the function y² = 4x. If x = 9 meters, find the rate of change of the balloon's height.</p>
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<p>A balloon rises such that its height is represented by the function y² = 4x. If x = 9 meters, find the rate of change of the balloon's height.</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 y² = 4x (height of the balloon)...(1)</p>
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<p>We have y² = 4x (height of the balloon)...(1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Take the derivative of y²: d/dx (y²) = 2y(dy/dx) = 4</p>
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<p>Take the derivative of y²: d/dx (y²) = 2y(dy/dx) = 4</p>
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<p>Given x = 9, we can find y: y² = 4(9) y = 6 (assuming y is positive)</p>
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<p>Given x = 9, we can find y: y² = 4(9) y = 6 (assuming y is positive)</p>
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<p>Substitute y = 6 into the derivative: 2(6)(dy/dx) = 4 12(dy/dx) = 4 dy/dx = 4/12 = 1/3</p>
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<p>Substitute y = 6 into the derivative: 2(6)(dy/dx) = 4 12(dy/dx) = 4 dy/dx = 4/12 = 1/3</p>
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<p>Hence, the rate of change of the balloon's height at x = 9 meters is 1/3.</p>
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<p>Hence, the rate of change of the balloon's height at x = 9 meters is 1/3.</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 by differentiating the function representing the balloon's height. We then solve for dy/dx using the given x value and simplify to find the final rate.</p>
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<p>We find the rate of change by differentiating the function representing the balloon's height. We then solve for dy/dx using the given x value and simplify to find the final 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 y² = x².</p>
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<p>Derive the second derivative of the function y² = 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>The first step is to find the first derivative, d/dx (y²) = 2y(dy/dx) = 2x...(1)</p>
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<p>The first step is to find the first derivative, d/dx (y²) = 2y(dy/dx) = 2x...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²/dx² (2y(dy/dx)) = d/dx (2x) d²/dx² (2y(dy/dx)) = 2</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²/dx² (2y(dy/dx)) = d/dx (2x) d²/dx² (2y(dy/dx)) = 2</p>
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<p>Using the product rule, d²/dx² (2y(dy/dx)) = 2[(dy/dx) + y(d²y/dx²)] 2[(dy/dx) + y(d²y/dx²)] = 2 (dy/dx) + y(d²y/dx²) = 1</p>
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<p>Using the product rule, d²/dx² (2y(dy/dx)) = 2[(dy/dx) + y(d²y/dx²)] 2[(dy/dx) + y(d²y/dx²)] = 2 (dy/dx) + y(d²y/dx²) = 1</p>
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<p>Therefore, the second derivative of the function y² = x² is (dy/dx) + y(d²y/dx²) = 1.</p>
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<p>Therefore, the second derivative of the function y² = x² is (dy/dx) + y(d²y/dx²) = 1.</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, starting with the first derivative. Using the product rule, we differentiate again to find the second derivative, then simplify the terms to find the final answer.</p>
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<p>We use the step-by-step process, starting with the first derivative. Using the product rule, we differentiate again to find the second derivative, then simplify the terms to find the final answer.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>Prove: d/dx (y³) = 3y²(dy/dx).</p>
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<p>Prove: d/dx (y³) = 3y²(dy/dx).</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:</p>
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<p>Let’s start using the chain rule:</p>
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<p>Consider y³ = (y)³</p>
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<p>Consider y³ = (y)³</p>
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<p>To differentiate, we use the chain rule: d/dx (y³) = 3y²(dy/dx)</p>
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<p>To differentiate, we use the chain rule: d/dx (y³) = 3y²(dy/dx)</p>
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<p>The derivative of the outer function is multiplied by the derivative of the inner function.</p>
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<p>The derivative of the outer function is multiplied by the derivative of the inner function.</p>
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<p>Hence, d/dx (y³) = 3y²(dy/dx) is proved.</p>
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<p>Hence, d/dx (y³) = 3y²(dy/dx) is 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. As a final step, we show the substitution of the derivative of the inner function into the equation to derive the final result.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. As a final step, we show the substitution of the derivative of the inner function into the equation to derive 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 5</h3>
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<h3>Problem 5</h3>
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<p>Solve: d/dx (y²/x)</p>
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<p>Solve: d/dx (y²/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 (y²/x) = (d/dx (y²) · x - y² · d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (y²/x) = (d/dx (y²) · x - y² · d/dx(x))/x²</p>
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<p>We will substitute d/dx (y²) = 2y(dy/dx) and d/dx (x) = 1 = (2y(dy/dx) · x - y² · 1)/x² = (2yx(dy/dx) - y²)/x²</p>
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<p>We will substitute d/dx (y²) = 2y(dy/dx) and d/dx (x) = 1 = (2y(dy/dx) · x - y² · 1)/x² = (2yx(dy/dx) - y²)/x²</p>
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<p>Therefore, d/dx (y²/x) = (2yx(dy/dx) - y²)/x²</p>
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<p>Therefore, d/dx (y²/x) = (2yx(dy/dx) - y²)/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 y²</h2>
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<h2>FAQs on the Derivative of y²</h2>
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<h3>1.Find the derivative of y².</h3>
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<h3>1.Find the derivative of y².</h3>
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<p>Using the chain rule to differentiate y² gives: d/dx (y²) = 2y(dy/dx) (simplified)</p>
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<p>Using the chain rule to differentiate y² gives: d/dx (y²) = 2y(dy/dx) (simplified)</p>
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<h3>2.Can we use the derivative of y² in real life?</h3>
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<h3>2.Can we use the derivative of y² in real life?</h3>
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<p>Yes, we can use the derivative of y² in real life to calculate rates of change in various fields such as physics, engineering, and biology.</p>
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<p>Yes, we can use the derivative of y² in real life to calculate rates of change in various fields such as physics, engineering, and biology.</p>
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<h3>3.Is it possible to take the derivative of y² at the point where y = 0?</h3>
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<h3>3.Is it possible to take the derivative of y² at the point where y = 0?</h3>
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<p>Yes, at y = 0, the derivative d/dx (y²) = 2y(dy/dx) = 0, since any constant multiple of 0 is 0.</p>
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<p>Yes, at y = 0, the derivative d/dx (y²) = 2y(dy/dx) = 0, since any constant multiple of 0 is 0.</p>
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<h3>4.What rule is used to differentiate y²/x?</h3>
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<h3>4.What rule is used to differentiate y²/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate y²/x: d/dx (y²/x) = (2yx(dy/dx) - y²)/x².</p>
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<p>We use the<a>quotient</a>rule to differentiate y²/x: d/dx (y²/x) = (2yx(dy/dx) - y²)/x².</p>
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<h3>5.Are the derivatives of y² and y³ the same?</h3>
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<h3>5.Are the derivatives of y² and y³ the same?</h3>
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<p>No, they are different. The derivative of y² is 2y(dy/dx), while the derivative of y³ is 3y²(dy/dx).</p>
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<p>No, they are different. The derivative of y² is 2y(dy/dx), while the derivative of y³ is 3y²(dy/dx).</p>
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<h2>Important Glossaries for the Derivative of y²</h2>
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<h2>Important Glossaries for the Derivative of y²</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating composite functions by differentiating the outer function and multiplying by the derivative of the inner function.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating composite functions by differentiating the outer function and multiplying by the derivative of the inner function.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule for differentiating products of two functions by adding the derivative of each function multiplied by the other function.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule for differentiating products of two functions by adding the derivative of each function multiplied by the other function.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating quotients of two functions by applying a specific formula involving the derivatives of the numerator and denominator.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating quotients of two functions by applying a specific formula involving the derivatives of the numerator and denominator.</li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>A derivative obtained by differentiating a function multiple times, indicating the rate of change of the rate of change.</li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>A derivative obtained by differentiating a function multiple times, indicating the rate of change of the rate of change.</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>