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2026-01-01
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<p>Last updated on<strong>September 20, 2025</strong></p>
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<p>Last updated on<strong>September 20, 2025</strong></p>
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<p>We use the derivative of y with respect to x as a measuring tool for how a function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of y with respect to x in detail.</p>
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<p>We use the derivative of y with respect to x as a measuring tool for how a function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of y with respect to x in detail.</p>
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<h2>What is the Derivative of y with respect to x?</h2>
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<h2>What is the Derivative of y with respect to x?</h2>
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<p>We now understand the derivative<a>of</a>y with respect to x. It is commonly represented as dy/dx and indicates how the<a>function</a>y changes as x changes. If a function y is differentiable within its domain, it has a clearly defined derivative. The key concepts are mentioned below:</p>
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<p>We now understand the derivative<a>of</a>y with respect to x. It is commonly represented as dy/dx and indicates how the<a>function</a>y changes as x changes. If a function y is differentiable within its domain, it has a clearly defined derivative. The key concepts are mentioned below:</p>
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<p><strong>Function:</strong>The relationship between x and y, where y is expressed as a function of x.</p>
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<p><strong>Function:</strong>The relationship between x and y, where y is expressed as a function of x.</p>
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<p><strong>Derivative:</strong>The measure of how a function changes as its input changes.</p>
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<p><strong>Derivative:</strong>The measure of how a function changes as its input changes.</p>
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<p><strong>Differentiability:</strong>A property indicating that a function has a derivative at every point in its domain.</p>
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<p><strong>Differentiability:</strong>A property indicating that a function has a derivative at every point in its domain.</p>
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<h2>Derivative of y with respect to x Formula</h2>
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<h2>Derivative of y with respect to x Formula</h2>
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<p>The derivative of y with respect to x can be denoted as dy/dx. If y is a function of x, the<a>formula</a>used to differentiate y with respect to x is based on the specific form of y.</p>
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<p>The derivative of y with respect to x can be denoted as dy/dx. If y is a function of x, the<a>formula</a>used to differentiate y with respect to x is based on the specific form of y.</p>
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<p>The formula applies to all x where the function y is continuous and differentiable.</p>
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<p>The formula applies to all x where the function y is continuous and differentiable.</p>
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<h2>Proofs of the Derivative of y with respect to x</h2>
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<h2>Proofs of the Derivative of y with respect to x</h2>
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<p>We can derive the derivative of y with respect to x using various proofs. To show this, we will use mathematical principles and rules of differentiation. There are several methods we use, such as:</p>
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<p>We can derive the derivative of y with respect to x using various proofs. To show this, we will use mathematical principles and rules of differentiation. There are several methods we use, such as:</p>
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<ol><li>By First Principle</li>
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<ol><li>By First Principle</li>
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<li>Using Chain Rule</li>
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<li>Using Chain Rule</li>
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<li>Using Product Rule</li>
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<li>Using Product Rule</li>
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</ol><p>We will now demonstrate the differentiation of y with respect to x using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate the differentiation of y with respect to x using the above-mentioned methods:</p>
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<h3>By First Principle</h3>
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<h3>By First Principle</h3>
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<p>The derivative of y with respect to x can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>The derivative of y with respect to x can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>To find the derivative of y using the first principle, we will consider f(x) = y. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h ... (1)</p>
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<p>To find the derivative of y using the first principle, we will consider f(x) = y. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h ... (1)</p>
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<p>Substituting y into<a>equation</a>(1), f'(x) = limₕ→₀ [y(x + h) - y(x)] / h = limₕ→₀ Δy / h where Δy = y(x + h) - y(x).</p>
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<p>Substituting y into<a>equation</a>(1), f'(x) = limₕ→₀ [y(x + h) - y(x)] / h = limₕ→₀ Δy / h where Δy = y(x + h) - y(x).</p>
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<h3>Using Chain Rule</h3>
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<h3>Using Chain Rule</h3>
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<p>To prove the differentiation of y using the chain rule, Consider y as a composite function. By chain rule, if y = g(f(x)), the derivative is dy/dx = g'(f(x))f'(x).</p>
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<p>To prove the differentiation of y using the chain rule, Consider y as a composite function. By chain rule, if y = g(f(x)), the derivative is dy/dx = g'(f(x))f'(x).</p>
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<h3>Using Product Rule</h3>
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<h3>Using Product Rule</h3>
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<p>We will now prove the derivative of a<a>product</a>of functions using the product rule. If y = u(x)v(x), then by product rule: dy/dx = u'(x)v(x) + u(x)v'(x).</p>
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<p>We will now prove the derivative of a<a>product</a>of functions using the product rule. If y = u(x)v(x), then by product rule: dy/dx = u'(x)v(x) + u(x)v'(x).</p>
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<h2>Higher-Order Derivatives of y with respect to x</h2>
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<h2>Higher-Order Derivatives of y with respect to x</h2>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky.</p>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky.</p>
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<p>To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like y(x).</p>
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<p>To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like y(x).</p>
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<p>For the first derivative of a function, we write dy/dx, 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 d²y/dx². Similarly, the third derivative, d³y/dx³ 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 dy/dx, 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 d²y/dx². Similarly, the third derivative, d³y/dx³ is the result of the second derivative and this pattern continues.</p>
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<p>For the nth derivative of y(x), we generally use dⁿy/dxⁿ, which tells us the change in the rate of change.</p>
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<p>For the nth derivative of y(x), we generally use dⁿy/dxⁿ, which tells us the change in the rate of change.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When the function y has a vertical asymptote at a certain point, the derivative is undefined at that point. When y is a<a>constant</a>function, the derivative dy/dx = 0.</p>
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<p>When the function y has a vertical asymptote at a certain point, the derivative is undefined at that point. When y is a<a>constant</a>function, the derivative dy/dx = 0.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of y with respect to x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of y with respect to x</h2>
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<p>Students frequently make mistakes when differentiating functions. 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 functions. 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 (y·y²)</p>
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<p>Calculate the derivative of (y·y²)</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) = y·y². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = y and v = y².</p>
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<p>Here, we have f(x) = y·y². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = y and v = y².</p>
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<p>Let’s differentiate each term, u′= dy/dx v′= d/dx (y²) = 2y(dy/dx)</p>
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<p>Let’s differentiate each term, u′= dy/dx v′= d/dx (y²) = 2y(dy/dx)</p>
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<p>substituting into the given equation, f'(x) = (dy/dx)·y² + y·2y(dy/dx)</p>
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<p>substituting into the given equation, f'(x) = (dy/dx)·y² + y·2y(dy/dx)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = y²(dy/dx) + 2y²(dy/dx).</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = y²(dy/dx) + 2y²(dy/dx).</p>
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<p>Thus, the derivative of the specified function is 3y²(dy/dx).</p>
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<p>Thus, the derivative of the specified function is 3y²(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 water tank is being filled with water at a rate that causes its height to change according to the function y = x², where y represents the height of the water in meters, and x is time in seconds. If x = 2 seconds, calculate the rate of change of the height of the water.</p>
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<p>A water tank is being filled with water at a rate that causes its height to change according to the function y = x², where y represents the height of the water in meters, and x is time in seconds. If x = 2 seconds, calculate the rate of change of the height of the water.</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 = x² (the height of the water)...(1)</p>
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<p>We have y = x² (the height of the water)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative of x²: dy/dx = 2x Given x = 2 (substitute this into the derivative) dy/dx = 2(2) = 4.</p>
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<p>Now, we will differentiate the equation (1) Take the derivative of x²: dy/dx = 2x Given x = 2 (substitute this into the derivative) dy/dx = 2(2) = 4.</p>
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<p>Hence, the rate of change of the height of the water at x = 2 seconds is 4 meters/second.</p>
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<p>Hence, the rate of change of the height of the water at x = 2 seconds is 4 meters/second.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the rate of change of the height of the water at x = 2 seconds as 4 meters/second, which means that at this time, the height of the water increases at a rate of 4 meters per second.</p>
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<p>We find the rate of change of the height of the water at x = 2 seconds as 4 meters/second, which means that at this time, the height of the water increases at a rate of 4 meters per second.</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 = x³.</p>
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<p>Derive the second derivative of the function y = x³.</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 = 3x²...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = 3x²...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [3x²] d²y/dx² = 6x.</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [3x²] d²y/dx² = 6x.</p>
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<p>Therefore, the second derivative of the function y = x³ is 6x.</p>
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<p>Therefore, the second derivative of the function y = x³ is 6x.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use the step-by-step process, where we start with the first derivative. By differentiating it, we find the second derivative, which is 6x.</p>
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<p>We use the step-by-step process, where we start with the first derivative. By differentiating it, we find the second derivative, which is 6x.</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 (x²) = 2x.</p>
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<p>Prove: d/dx (x²) = 2x.</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 = x². To differentiate, we use the power rule: dy/dx = 2x. Hence proved.</p>
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<p>Consider y = x². To differentiate, we use the power rule: dy/dx = 2x. 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 power rule to differentiate the equation. We directly obtain the derivative as 2x.</p>
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<p>In this step-by-step process, we used the power rule to differentiate the equation. We directly obtain the derivative as 2x.</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 (y/x)</p>
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<p>Solve: d/dx (y/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 (y/x) = (d/dx (y)·x - y·d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (y/x) = (d/dx (y)·x - y·d/dx(x))/x²</p>
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<p>We will substitute d/dx (y) = dy/dx and d/dx (x) = 1 = (x(dy/dx) - y)/x².</p>
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<p>We will substitute d/dx (y) = dy/dx and d/dx (x) = 1 = (x(dy/dx) - y)/x².</p>
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<p>Therefore, d/dx (y/x) = (x(dy/dx) - y)/x².</p>
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<p>Therefore, d/dx (y/x) = (x(dy/dx) - y)/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 y with respect to x</h2>
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<h2>FAQs on the Derivative of y with respect to x</h2>
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<h3>1.Find the derivative of y = x².</h3>
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<h3>1.Find the derivative of y = x².</h3>
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<p>Using the<a>power</a>rule for y = x², dy/dx = 2x.</p>
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<p>Using the<a>power</a>rule for y = x², dy/dx = 2x.</p>
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<h3>2.Can we use the derivative of y with respect to x in real life?</h3>
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<h3>2.Can we use the derivative of y with respect to x in real life?</h3>
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<p>Yes, we can use the derivative of y with respect to x in real life to calculate the rate of change of any motion, especially in fields such as mathematics, physics, and economics.</p>
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<p>Yes, we can use the derivative of y with respect to x in real life to calculate the rate of change of any motion, especially in fields such as mathematics, physics, and economics.</p>
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<h3>3.Is it possible to take the derivative of y at a point where the function is not defined?</h3>
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<h3>3.Is it possible to take the derivative of y at a point where the function is not defined?</h3>
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<p>No, at points where the function is not defined, it is impossible to take the derivative (since the function does not exist there).</p>
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<p>No, at points where the function is not defined, it is impossible to take the derivative (since the function does not exist there).</p>
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<h3>4.What rule is used to differentiate y/x?</h3>
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<h3>4.What rule is used to differentiate y/x?</h3>
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<p>We use the quotient rule to differentiate y/x, d/dx (y/x) = (x(dy/dx) - y)/x².</p>
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<p>We use the quotient rule to differentiate y/x, d/dx (y/x) = (x(dy/dx) - y)/x².</p>
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<h3>5.Are the derivatives of y and y⁻¹ the same?</h3>
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<h3>5.Are the derivatives of y and y⁻¹ the same?</h3>
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<p>No, they are different. The derivative of y = x is 1, while the derivative of y⁻¹ is -1/y².</p>
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<p>No, they are different. The derivative of y = x is 1, while the derivative of y⁻¹ is -1/y².</p>
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<h2>Important Glossaries for the Derivative of y with respect to x</h2>
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<h2>Important Glossaries for the Derivative of y with respect to x</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>Continuous Function:</strong>A function that is continuous at every point in its domain.</li>
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</ul><ul><li><strong>Continuous Function:</strong>A function that is continuous at every point in its domain.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate expressions in the form of a quotient.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate expressions in the form of a quotient.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate composite functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate composite functions.</li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>Derivatives obtained by differentiating a function multiple times.</li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>Derivatives obtained by differentiating a function multiple times.</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>