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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 x^-1/2, which is -1/2 * x^-3/2, as a tool to understand how the function changes in response to a slight change in x. Derivatives help us calculate various rates of change in real-life situations. We will now talk about the derivative of x^-1/2 in detail.</p>
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<p>We use the derivative of x^-1/2, which is -1/2 * x^-3/2, as a tool to understand how the function changes in response to a slight change in x. Derivatives help us calculate various rates of change in real-life situations. We will now talk about the derivative of x^-1/2 in detail.</p>
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<h2>What is the Derivative of x^-1/2?</h2>
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<h2>What is the Derivative of x^-1/2?</h2>
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<p>We now understand the derivative of x^-1/2. It is commonly represented as d/dx (x^-1/2) or (x^-1/2)', and its value is -1/2 * x^-3/2.</p>
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<p>We now understand the derivative of x^-1/2. It is commonly represented as d/dx (x^-1/2) or (x^-1/2)', and its value is -1/2 * x^-3/2.</p>
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<p>The<a>function</a>x^-1/2 has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
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<p>The<a>function</a>x^-1/2 has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
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<p>Power Function: (x^n where n = -1/2).</p>
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<p>Power Function: (x^n where n = -1/2).</p>
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<p>Power Rule: Rule for differentiating x^n.</p>
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<p>Power Rule: Rule for differentiating x^n.</p>
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<p>Negative Exponent: A<a>negative exponent</a>indicates the reciprocal of the<a>base</a>raised to the<a>absolute value</a>of the exponent.</p>
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<p>Negative Exponent: A<a>negative exponent</a>indicates the reciprocal of the<a>base</a>raised to the<a>absolute value</a>of the exponent.</p>
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<h2>Derivative of x^-1/2 Formula</h2>
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<h2>Derivative of x^-1/2 Formula</h2>
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<p>The derivative of x^-1/2 can be denoted as d/dx (x^-1/2) or (x^-1/2)'.</p>
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<p>The derivative of x^-1/2 can be denoted as d/dx (x^-1/2) or (x^-1/2)'.</p>
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<p>The<a>formula</a>we use to differentiate x^-1/2 is: d/dx (x^-1/2) = -1/2 * x^-3/2</p>
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<p>The<a>formula</a>we use to differentiate x^-1/2 is: d/dx (x^-1/2) = -1/2 * x^-3/2</p>
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<p>The formula applies to all x where x > 0.</p>
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<p>The formula applies to all x where x > 0.</p>
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<h2>Proofs of the Derivative of x^-1/2</h2>
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<h2>Proofs of the Derivative of x^-1/2</h2>
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<p>We can derive the derivative of x^-1/2 using proofs. To show this, we will use the<a>power</a>rule of differentiation. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of x^-1/2 using proofs. To show this, we will use the<a>power</a>rule of differentiation. There are several methods we use to prove this, 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 Power Rule</li>
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<li>Using Power Rule</li>
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</ol><p>We will now demonstrate that the differentiation of x^-1/2 results in -1/2 * x^-3/2 using these methods:</p>
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</ol><p>We will now demonstrate that the differentiation of x^-1/2 results in -1/2 * x^-3/2 using these 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 x^-1/2 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 x^-1/2 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 x^-1/2 using the first principle, we will consider f(x) = x^-1/2. 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 x^-1/2 using the first principle, we will consider f(x) = x^-1/2. 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>Given that f(x) = x^-1/2, we write f(x + h) = (x + h)^-1/2.</p>
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<p>Given that f(x) = x^-1/2, we write f(x + h) = (x + h)^-1/2.</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [(x + h)^-1/2 - x^-1/2] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [(x + h)^-1/2 - x^-1/2] / h</p>
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<p>Using algebraic manipulation to combine the<a>fractions</a>, = limₕ→₀ [(x^-1/2(1 - (x + h)^1/2/x^1/2))] / h</p>
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<p>Using algebraic manipulation to combine the<a>fractions</a>, = limₕ→₀ [(x^-1/2(1 - (x + h)^1/2/x^1/2))] / h</p>
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<p>As h approaches zero, we get the derivative, f'(x) = -1/2 * x^-3/2.</p>
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<p>As h approaches zero, we get the derivative, f'(x) = -1/2 * x^-3/2.</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h3>Using Power Rule</h3>
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<h3>Using Power Rule</h3>
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<p>To prove the differentiation of x^-1/2 using the power rule, We use the formula: d/dx (x^n) = n * x^(n-1).</p>
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<p>To prove the differentiation of x^-1/2 using the power rule, We use the formula: d/dx (x^n) = n * x^(n-1).</p>
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<p>Substituting n = -1/2 in the power rule, d/dx (x^-1/2) = (-1/2) * x^(-1/2 - 1) = -1/2 * x^-3/2.</p>
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<p>Substituting n = -1/2 in the power rule, d/dx (x^-1/2) = (-1/2) * x^(-1/2 - 1) = -1/2 * x^-3/2.</p>
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<p>Thus, the derivative of x^-1/2 is -1/2 * x^-3/2.</p>
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<p>Thus, the derivative of x^-1/2 is -1/2 * x^-3/2.</p>
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<h2>Higher-Order Derivatives of x^-1/2</h2>
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<h2>Higher-Order Derivatives of x^-1/2</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 x^-1/2.</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 x^-1/2.</p>
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<p>For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x) is the result of the second derivative and this pattern continues.</p>
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<p>For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x) is the result of the second derivative and this pattern continues.</p>
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<p>For the nth derivative of x^-1/2, we generally use f^(n)(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 x^-1/2, we generally use f^(n)(x) for the nth derivative of a function f(x) which tells us the change in the rate of change (continuing for higher-order derivatives).</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When x is 0, the derivative is undefined because x^-1/2 has a vertical asymptote there. When x is 1, the derivative of x^-1/2 = -1/2 * 1^-3/2, which is -1/2.</p>
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<p>When x is 0, the derivative is undefined because x^-1/2 has a vertical asymptote there. When x is 1, the derivative of x^-1/2 = -1/2 * 1^-3/2, which is -1/2.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of x^-1/2</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of x^-1/2</h2>
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<p>Students frequently make mistakes when differentiating x^-1/2. 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 x^-1/2. 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 (x^-1/2 * x^3).</p>
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<p>Calculate the derivative of (x^-1/2 * x^3).</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) = x^-1/2 * x^3.</p>
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<p>Here, we have f(x) = x^-1/2 * x^3.</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x^-1/2 and v = x^3.</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x^-1/2 and v = x^3.</p>
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<p>Let’s differentiate each term, u′ = d/dx (x^-1/2) = -1/2 * x^-3/2 v′ = d/dx (x^3) = 3x^2</p>
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<p>Let’s differentiate each term, u′ = d/dx (x^-1/2) = -1/2 * x^-3/2 v′ = d/dx (x^3) = 3x^2</p>
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<p>Substituting into the given equation, f'(x) = (-1/2 * x^-3/2) * x^3 + (x^-1/2) * 3x^2</p>
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<p>Substituting into the given equation, f'(x) = (-1/2 * x^-3/2) * x^3 + (x^-1/2) * 3x^2</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -1/2 * x^3/2 + 3x^3/2</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -1/2 * x^3/2 + 3x^3/2</p>
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<p>Thus, the derivative of the specified function is (5/2) * x^3/2.</p>
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<p>Thus, the derivative of the specified function is (5/2) * x^3/2.</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 chemical concentration decreases over time according to the function C(t) = t^-1/2. Calculate the rate of change of concentration at t = 4 hours.</p>
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<p>A chemical concentration decreases over time according to the function C(t) = t^-1/2. Calculate the rate of change of concentration at t = 4 hours.</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 C(t) = t^-1/2 (concentration function)...(1)</p>
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<p>We have C(t) = t^-1/2 (concentration function)...(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 t^-1/2: dC/dt = -1/2 * t^-3/2</p>
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<p>Take the derivative of t^-1/2: dC/dt = -1/2 * t^-3/2</p>
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<p>Given t = 4 (substitute this into the derivative) dC/dt = -1/2 * (4)^-3/2 = -1/2 * 1/8 = -1/16</p>
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<p>Given t = 4 (substitute this into the derivative) dC/dt = -1/2 * (4)^-3/2 = -1/2 * 1/8 = -1/16</p>
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<p>Hence, the rate of change of concentration at t = 4 hours is -1/16.</p>
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<p>Hence, the rate of change of concentration at t = 4 hours is -1/16.</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 concentration at t = 4 hours as -1/16, which means the concentration is decreasing at that rate at that specific time.</p>
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<p>We find the rate of change of concentration at t = 4 hours as -1/16, which means the concentration is decreasing at that rate at that specific time.</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 = x^-1/2.</p>
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<p>Derive the second derivative of the function y = x^-1/2.</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 = -1/2 * x^-3/2...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = -1/2 * x^-3/2...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/2 * x^-3/2] = -1/2 * (-3/2 * x^-5/2) = 3/4 * x^-5/2</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/2 * x^-3/2] = -1/2 * (-3/2 * x^-5/2) = 3/4 * x^-5/2</p>
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<p>Therefore, the second derivative of the function y = x^-1/2 is 3/4 * x^-5/2.</p>
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<p>Therefore, the second derivative of the function y = x^-1/2 is 3/4 * x^-5/2.</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. Using the power rule, we differentiate -1/2 * x^-3/2. We then simplify the terms to find the final answer.</p>
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<p>We use the step-by-step process, where we start with the first derivative. Using the power rule, we differentiate -1/2 * x^-3/2. We 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 ((3x)^-1/2) = -3/2 * (3x)^-3/2.</p>
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<p>Prove: d/dx ((3x)^-1/2) = -3/2 * (3x)^-3/2.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Let’s start using the chain rule: Consider y = (3x)^-1/2</p>
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<p>Let’s start using the chain rule: Consider y = (3x)^-1/2</p>
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<p>To differentiate, we use the chain rule: dy/dx = -1/2 * (3x)^-3/2 * d/dx(3x)</p>
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<p>To differentiate, we use the chain rule: dy/dx = -1/2 * (3x)^-3/2 * d/dx(3x)</p>
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<p>Since the derivative of 3x is 3, dy/dx = -1/2 * (3x)^-3/2 * 3 = -3/2 * (3x)^-3/2</p>
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<p>Since the derivative of 3x is 3, dy/dx = -1/2 * (3x)^-3/2 * 3 = -3/2 * (3x)^-3/2</p>
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<p>Hence proved.</p>
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<p>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 replace 3x with its derivative. As a final step, we substitute y = (3x)^-1/2 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 replace 3x with its derivative. As a final step, we substitute y = (3x)^-1/2 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 ((x^2)/x^-1/2).</p>
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<p>Solve: d/dx ((x^2)/x^-1/2).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>To differentiate the function, we use the quotient rule: d/dx ((x^2)/x^-1/2) = (d/dx (x^2) * x^-1/2 - x^2 * d/dx (x^-1/2)) / (x^-1/2)^2</p>
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<p>To differentiate the function, we use the quotient rule: d/dx ((x^2)/x^-1/2) = (d/dx (x^2) * x^-1/2 - x^2 * d/dx (x^-1/2)) / (x^-1/2)^2</p>
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<p>We will substitute d/dx (x^2) = 2x and d/dx (x^-1/2) = -1/2 * x^-3/2 = (2x * x^-1/2 - x^2 * (-1/2 * x^-3/2)) / (x^-1) = (2x^1/2 + 1/2 * x^3/2) / (x^-1)</p>
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<p>We will substitute d/dx (x^2) = 2x and d/dx (x^-1/2) = -1/2 * x^-3/2 = (2x * x^-1/2 - x^2 * (-1/2 * x^-3/2)) / (x^-1) = (2x^1/2 + 1/2 * x^3/2) / (x^-1)</p>
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<p>Therefore, d/dx ((x^2)/x^-1/2) = (5/2) * x^1/2.</p>
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<p>Therefore, d/dx ((x^2)/x^-1/2) = (5/2) * x^1/2.</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 x^-1/2</h2>
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<h2>FAQs on the Derivative of x^-1/2</h2>
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<h3>1.Find the derivative of x^-1/2.</h3>
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<h3>1.Find the derivative of x^-1/2.</h3>
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<p>Using the power rule to differentiate x^-1/2, d/dx (x^-1/2) = -1/2 * x^-3/2.</p>
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<p>Using the power rule to differentiate x^-1/2, d/dx (x^-1/2) = -1/2 * x^-3/2.</p>
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<h3>2.Can we use the derivative of x^-1/2 in real life?</h3>
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<h3>2.Can we use the derivative of x^-1/2 in real life?</h3>
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<p>Yes, we can use the derivative of x^-1/2 in real life in calculating rates of change, especially in fields such as physics and engineering where inverse<a>square</a>laws are common.</p>
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<p>Yes, we can use the derivative of x^-1/2 in real life in calculating rates of change, especially in fields such as physics and engineering where inverse<a>square</a>laws are common.</p>
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<h3>3.Is it possible to take the derivative of x^-1/2 at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of x^-1/2 at the point where x = 0?</h3>
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<p>No, x = 0 is a point where x^-1/2 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, x = 0 is a point where x^-1/2 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 x^-1/2 * x?</h3>
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<h3>4.What rule is used to differentiate x^-1/2 * x?</h3>
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<p>We use the<a>product</a>rule to differentiate x^-1/2 * x, d/dx (x^-1/2 * x) = (x^-1/2) * d/dx (x) + x * d/dx (x^-1/2).</p>
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<p>We use the<a>product</a>rule to differentiate x^-1/2 * x, d/dx (x^-1/2 * x) = (x^-1/2) * d/dx (x) + x * d/dx (x^-1/2).</p>
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<h3>5.Are the derivatives of x^-1/2 and (x^-1/2)^-1 the same?</h3>
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<h3>5.Are the derivatives of x^-1/2 and (x^-1/2)^-1 the same?</h3>
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<p>No, they are different. The derivative of x^-1/2 is -1/2 * x^-3/2, while the derivative of (x^-1/2)^-1, which is x^1/2, is 1/2 * x^-1/2.</p>
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<p>No, they are different. The derivative of x^-1/2 is -1/2 * x^-3/2, while the derivative of (x^-1/2)^-1, which is x^1/2, is 1/2 * x^-1/2.</p>
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<h2>Important Glossaries for the Derivative of x^-1/2</h2>
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<h2>Important Glossaries for the Derivative of x^-1/2</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>Power Rule:</strong>A differentiation rule used for functions of the form x^n.</li>
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</ul><ul><li><strong>Power Rule:</strong>A differentiation rule used for functions of the form x^n.</li>
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</ul><ul><li><strong>Negative Exponent:</strong>An exponent that indicates the reciprocal of the base raised to the absolute value of the exponent.</li>
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</ul><ul><li><strong>Negative Exponent:</strong>An exponent that indicates the reciprocal of the base raised to the absolute value of the exponent.</li>
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</ul><ul><li><strong>Reciprocal:</strong>A value that, when multiplied by the original number, results in 1.</li>
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</ul><ul><li><strong>Reciprocal:</strong>A value that, when multiplied by the original number, results in 1.</li>
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</ul><ul><li><strong>Asymptote:</strong>A line that a curve approaches without crossing.</li>
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</ul><ul><li><strong>Asymptote:</strong>A line that a curve approaches without crossing.</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>