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<p>Last updated on<strong>October 10, 2025</strong></p>
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<p>Last updated on<strong>October 10, 2025</strong></p>
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<p>We use the derivative of a parabola, typically represented as a quadratic function, as a tool to measure how the function's rate of change varies with respect to x. Derivatives help us calculate rates, such as acceleration, in real-life situations. We will now explore the derivative of a parabola in detail.</p>
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<p>We use the derivative of a parabola, typically represented as a quadratic function, as a tool to measure how the function's rate of change varies with respect to x. Derivatives help us calculate rates, such as acceleration, in real-life situations. We will now explore the derivative of a parabola in detail.</p>
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<h2>What is the Derivative of a Parabola?</h2>
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<h2>What is the Derivative of a Parabola?</h2>
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<p>We now understand the derivative of a parabola. It is commonly represented as d/dx (ax² + bx + c) or (ax² + bx + c)', and its value is 2ax + b. The quadratic<a>function</a>ax² + bx + c has a clearly defined derivative, indicating it is differentiable across its domain.</p>
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<p>We now understand the derivative of a parabola. It is commonly represented as d/dx (ax² + bx + c) or (ax² + bx + c)', and its value is 2ax + b. The quadratic<a>function</a>ax² + bx + c has a clearly defined derivative, indicating it is differentiable across 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>Quadratic Function: A function of the form ax² + bx + c.</p>
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<p>Quadratic Function: A function of the form ax² + bx + c.</p>
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<p>Power Rule: A rule for differentiating<a>terms</a>like ax².</p>
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<p>Power Rule: A rule for differentiating<a>terms</a>like ax².</p>
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<p>Linear Function: The derivative of a quadratic function is a linear function.</p>
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<p>Linear Function: The derivative of a quadratic function is a linear function.</p>
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<h2>Derivative of Parabola Formula</h2>
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<h2>Derivative of Parabola Formula</h2>
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<p>The derivative of a parabola can be denoted as d/dx (ax² + bx + c) or (ax² + bx + c)'.</p>
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<p>The derivative of a parabola can be denoted as d/dx (ax² + bx + c) or (ax² + bx + c)'.</p>
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<p>The<a>formula</a>we use to differentiate a quadratic function is: d/dx (ax² + bx + c) = 2ax + b</p>
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<p>The<a>formula</a>we use to differentiate a quadratic function is: d/dx (ax² + bx + c) = 2ax + b</p>
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<p>The formula applies to all x within the domain of a parabola.</p>
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<p>The formula applies to all x within the domain of a parabola.</p>
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<h2>Proofs of the Derivative of a Parabola</h2>
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<h2>Proofs of the Derivative of a Parabola</h2>
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<p>We can derive the derivative of a parabola using proofs. To show this, we will use<a>algebraic identities</a>along with the rules of differentiation.</p>
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<p>We can derive the derivative of a parabola using proofs. To show this, we will use<a>algebraic identities</a>along with the rules of differentiation.</p>
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<p>There are several methods we use to prove this, such as:</p>
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<p>There are several methods we use to prove this, such as:</p>
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<ul><li>By First Principle </li>
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<ul><li>By First Principle </li>
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<li>Using the Power Rule</li>
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<li>Using the Power Rule</li>
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</ul><p>We will now demonstrate that the differentiation of ax² + bx + c results in 2ax + b using the above-mentioned methods:</p>
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</ul><p>We will now demonstrate that the differentiation of ax² + bx + c results in 2ax + b using the above-mentioned methods:</p>
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<h2>By First Principle</h2>
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<h2>By First Principle</h2>
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<p>The derivative of ax² + bx + c can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative using the first principle, we will consider f(x) = ax² + bx + c. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = ax² + bx + c, we write f(x + h) = a(x + h)² + b(x + h) + c. Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [a(x + h)² + b(x + h) + c - (ax² + bx + c)] / h = limₕ→₀ [a(x² + 2xh + h²) + bx + bh + c - ax² - bx - c] / h = limₕ→₀ [2axh + ah² + bh] / h = limₕ→₀ [2ax + ah + b] As h approaches 0, the term ah vanishes, f'(x) = 2ax + b.</p>
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<p>The derivative of ax² + bx + c can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative using the first principle, we will consider f(x) = ax² + bx + c. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = ax² + bx + c, we write f(x + h) = a(x + h)² + b(x + h) + c. Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [a(x + h)² + b(x + h) + c - (ax² + bx + c)] / h = limₕ→₀ [a(x² + 2xh + h²) + bx + bh + c - ax² - bx - c] / h = limₕ→₀ [2axh + ah² + bh] / h = limₕ→₀ [2ax + ah + b] As h approaches 0, the term ah vanishes, f'(x) = 2ax + b.</p>
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<h2>Using the Power Rule</h2>
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<h2>Using the Power Rule</h2>
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<p>To prove the differentiation of ax² + bx + c using the<a>power</a>rule, We differentiate each term separately: d/dx (ax²) = 2ax d/dx (bx) = b d/dx (c) = 0 Combining these gives: d/dx (ax² + bx + c) = 2ax + b.</p>
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<p>To prove the differentiation of ax² + bx + c using the<a>power</a>rule, We differentiate each term separately: d/dx (ax²) = 2ax d/dx (bx) = b d/dx (c) = 0 Combining these gives: d/dx (ax² + bx + c) = 2ax + b.</p>
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<h2>Higher-Order Derivatives of a Parabola</h2>
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<h2>Higher-Order Derivatives of a Parabola</h2>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. For a quadratic function, the first derivative is linear, and the second derivative is<a>constant</a>. To understand them better, consider a car where the position changes (function), the speed changes (first derivative), and the acceleration (second derivative) is constant. Higher-order derivatives simplify<a>understanding of</a>functions like ax² + bx + c.</p>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. For a quadratic function, the first derivative is linear, and the second derivative is<a>constant</a>. To understand them better, consider a car where the position changes (function), the speed changes (first derivative), and the acceleration (second derivative) is constant. Higher-order derivatives simplify<a>understanding of</a>functions like ax² + bx + c.</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), and is constant for a parabola.</p>
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<p>The second derivative is derived from the first derivative, which is denoted using f′′(x), and is constant for a parabola.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When the<a>coefficient</a>a is zero, the derivative reduces to a constant value b, representing a linear function. If b is also zero, the derivative becomes zero, indicating a constant function.</p>
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<p>When the<a>coefficient</a>a is zero, the derivative reduces to a constant value b, representing a linear function. If b is also zero, the derivative becomes zero, indicating a constant function.</p>
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<p>If a is positive, the parabola opens upwards, and its derivative increases. If a is negative, the parabola opens downwards, and its derivative decreases.</p>
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<p>If a is positive, the parabola opens upwards, and its derivative increases. If a is negative, the parabola opens downwards, and its derivative decreases.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of a Parabola</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of a Parabola</h2>
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<p>Students frequently make mistakes when differentiating quadratic 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 quadratic 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 the function f(x) = 3x² + 4x + 5.</p>
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<p>Calculate the derivative of the function f(x) = 3x² + 4x + 5.</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) = 3x² + 4x + 5. Using the power rule, f'(x) = d/dx (3x²) + d/dx (4x) + d/dx (5) = 6x + 4 + 0 Thus, the derivative of the specified function is 6x + 4.</p>
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<p>Here, we have f(x) = 3x² + 4x + 5. Using the power rule, f'(x) = d/dx (3x²) + d/dx (4x) + d/dx (5) = 6x + 4 + 0 Thus, the derivative of the specified function is 6x + 4.</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 differentiating each term separately using the power rule and then combining the results to get the final derivative.</p>
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<p>We find the derivative of the given function by differentiating each term separately using the power rule and then combining the results to get the final derivative.</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 ball is thrown vertically upwards, and its height as a function of time is given by h(t) = -5t² + 20t + 3. Calculate the velocity of the ball at t = 2 seconds.</p>
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<p>A ball is thrown vertically upwards, and its height as a function of time is given by h(t) = -5t² + 20t + 3. Calculate the velocity of the ball at t = 2 seconds.</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 h(t) = -5t² + 20t + 3 (height of the ball)... Now, we will differentiate h(t) to find velocity, v(t): v(t) = d/dt (-5t² + 20t + 3) v(t) = -10t + 20 Substitute t = 2 seconds into the velocity function: v(2) = -10(2) + 20 = -20 + 20 = 0 Hence, the velocity of the ball at t = 2 seconds is 0 m/s, indicating it momentarily stops.</p>
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<p>We have h(t) = -5t² + 20t + 3 (height of the ball)... Now, we will differentiate h(t) to find velocity, v(t): v(t) = d/dt (-5t² + 20t + 3) v(t) = -10t + 20 Substitute t = 2 seconds into the velocity function: v(2) = -10(2) + 20 = -20 + 20 = 0 Hence, the velocity of the ball at t = 2 seconds is 0 m/s, indicating it momentarily stops.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the velocity of the ball at t = 2 seconds by differentiating the height function to get the velocity function and then substituting the given time into the velocity function.</p>
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<p>We find the velocity of the ball at t = 2 seconds by differentiating the height function to get the velocity function and then substituting the given time into the velocity function.</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 = 2x² + 3x + 1.</p>
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<p>Derive the second derivative of the function y = 2x² + 3x + 1.</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 = d/dx (2x² + 3x + 1) = 4x + 3 Now, we differentiate again to get the second derivative: d²y/dx² = d/dx (4x + 3) = 4 Therefore, the second derivative of the function y = 2x² + 3x + 1 is 4.</p>
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<p>The first step is to find the first derivative, dy/dx = d/dx (2x² + 3x + 1) = 4x + 3 Now, we differentiate again to get the second derivative: d²y/dx² = d/dx (4x + 3) = 4 Therefore, the second derivative of the function y = 2x² + 3x + 1 is 4.</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.</p>
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<p>We use the step-by-step process, starting with the first derivative.</p>
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<p>Then, we differentiate the result to find the constant second derivative.</p>
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<p>Then, we differentiate the result to find the constant 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 (3x²) = 6x.</p>
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<p>Prove: d/dx (3x²) = 6x.</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 power rule: Consider y = 3x² To differentiate, we apply the power rule: dy/dx = 2·3x^(2-1) = 6x Hence, proved.</p>
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<p>Let's start using the power rule: Consider y = 3x² To differentiate, we apply the power rule: dy/dx = 2·3x^(2-1) = 6x 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 by multiplying the coefficient by the power, reducing the power by one.</p>
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<p>In this step-by-step process, we used the power rule to differentiate the equation by multiplying the coefficient by the power, reducing the power by one.</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 (x² + 2x + 1/x).</p>
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<p>Solve: d/dx (x² + 2x + 1/x).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>To differentiate the function, we use the power rule and the rule for differentiating x⁻¹: d/dx (x² + 2x + 1/x) = d/dx (x²) + d/dx (2x) + d/dx (x⁻¹) = 2x + 2 - x⁻² Therefore, d/dx (x² + 2x + 1/x) = 2x + 2 - 1/x².</p>
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<p>To differentiate the function, we use the power rule and the rule for differentiating x⁻¹: d/dx (x² + 2x + 1/x) = d/dx (x²) + d/dx (2x) + d/dx (x⁻¹) = 2x + 2 - x⁻² Therefore, d/dx (x² + 2x + 1/x) = 2x + 2 - 1/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 each term using the power rule and the derivative of x⁻¹.</p>
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<p>In this process, we differentiate each term using the power rule and the derivative of x⁻¹.</p>
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<p>We then simplify the expression to obtain the final result.</p>
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<p>We then simplify the expression 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 a Parabola</h2>
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<h2>FAQs on the Derivative of a Parabola</h2>
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<h3>1.Find the derivative of 4x² + 6x + 9.</h3>
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<h3>1.Find the derivative of 4x² + 6x + 9.</h3>
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<p>Using the power rule, d/dx (4x² + 6x + 9) = 8x + 6.</p>
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<p>Using the power rule, d/dx (4x² + 6x + 9) = 8x + 6.</p>
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<h3>2.Can we use the derivative of a parabola in real life?</h3>
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<h3>2.Can we use the derivative of a parabola in real life?</h3>
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<p>Yes, derivatives of parabolas are used in real life to calculate rates of change, such as velocity and acceleration, especially in physics and engineering.</p>
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<p>Yes, derivatives of parabolas are used in real life to calculate rates of change, such as velocity and acceleration, especially in physics and engineering.</p>
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<h3>3.What is the second derivative of a parabola?</h3>
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<h3>3.What is the second derivative of a parabola?</h3>
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<p>The second derivative of a parabola ax² + bx + c is constant, equal to 2a.</p>
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<p>The second derivative of a parabola ax² + bx + c is constant, equal to 2a.</p>
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<h3>4.What rule is used to differentiate a quadratic function?</h3>
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<h3>4.What rule is used to differentiate a quadratic function?</h3>
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<p>We use the power rule to differentiate each term of a quadratic function ax² + bx + c.</p>
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<p>We use the power rule to differentiate each term of a quadratic function ax² + bx + c.</p>
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<h3>5.Are the derivatives of ax² + bx + c and (ax² + bx + c)⁻¹ the same?</h3>
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<h3>5.Are the derivatives of ax² + bx + c and (ax² + bx + c)⁻¹ the same?</h3>
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<p>No, they are different. The derivative of ax² + bx + c is 2ax + b, while the derivative of (ax² + bx + c)⁻¹ involves the chain rule.</p>
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<p>No, they are different. The derivative of ax² + bx + c is 2ax + b, while the derivative of (ax² + bx + c)⁻¹ involves the chain rule.</p>
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<h2>Important Glossaries for the Derivative of a Parabola</h2>
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<h2>Important Glossaries for the Derivative of a Parabola</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates the rate of change of the function with respect to a variable.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates the rate of change of the function with respect to a variable.</li>
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</ul><ul><li><strong>Quadratic Function:</strong>A second-degree polynomial function of the form ax² + bx + c.</li>
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</ul><ul><li><strong>Quadratic Function:</strong>A second-degree polynomial function of the form ax² + bx + c.</li>
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</ul><ul><li><strong>Power Rule:</strong>A fundamental rule in calculus for differentiating functions of the form xⁿ.</li>
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</ul><ul><li><strong>Power Rule:</strong>A fundamental rule in calculus for differentiating functions of the form xⁿ.</li>
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</ul><ul><li><strong>First Derivative:</strong>The initial derivative of a function, representing the rate of change or slope of the function.</li>
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</ul><ul><li><strong>First Derivative:</strong>The initial derivative of a function, representing the rate of change or slope of the function.</li>
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</ul><ul><li><strong>Second Derivative:</strong>The derivative of the first derivative, indicating the rate of change of the rate of change, often related to acceleration.</li>
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</ul><ul><li><strong>Second Derivative:</strong>The derivative of the first derivative, indicating the rate of change of the rate of change, often related to acceleration.</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>