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2026-01-01
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<p>Last updated on<strong>October 8, 2025</strong></p>
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<p>Last updated on<strong>October 8, 2025</strong></p>
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<p>We use the derivative of x^(1/x) to understand how this function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of x^(1/x) in detail.</p>
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<p>We use the derivative of x^(1/x) to understand how this function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of x^(1/x) in detail.</p>
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<h2>What is the Derivative of x^(1/x)?</h2>
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<h2>What is the Derivative of x^(1/x)?</h2>
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<p>We now understand the derivative<a>of</a>x(1/x). It is commonly represented as d/dx (x(1/x)) or (x(1/x))', and its value is a bit more complex. The<a>function</a>x(1/x) has a defined derivative, indicating it is differentiable within its domain.</p>
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<p>We now understand the derivative<a>of</a>x(1/x). It is commonly represented as d/dx (x(1/x)) or (x(1/x))', and its value is a bit more complex. The<a>function</a>x(1/x) has a 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>Power Function: x(1/x) can be rewritten as e(ln(x)/x).</p>
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<p>Power Function: x(1/x) can be rewritten as e(ln(x)/x).</p>
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<p>Chain Rule: Rule for differentiating functions like e(ln(x)/x).</p>
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<p>Chain Rule: Rule for differentiating functions like e(ln(x)/x).</p>
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<p>Logarithmic Differentiation: Useful for differentiating functions like x(1/x).</p>
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<p>Logarithmic Differentiation: Useful for differentiating functions like x(1/x).</p>
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<h2>Derivative of x^(1/x) Formula</h2>
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<h2>Derivative of x^(1/x) Formula</h2>
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<p>The derivative of x(1/x) can be denoted as d/dx (x(1/x)) or (x(1/x))'.</p>
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<p>The derivative of x(1/x) can be denoted as d/dx (x(1/x)) or (x(1/x))'.</p>
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<p>The<a>formula</a>we use to differentiate x(1/x) involves logarithmic differentiation: d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²]</p>
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<p>The<a>formula</a>we use to differentiate x(1/x) involves logarithmic differentiation: d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²]</p>
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<p>This formula applies to all x > 0.</p>
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<p>This formula applies to all x > 0.</p>
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<h2>Proofs of the Derivative of x^(1/x)</h2>
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<h2>Proofs of the Derivative of x^(1/x)</h2>
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<p>We can derive the derivative of x(1/x) using proofs. To show this, we will use logarithmic differentiation along with the rules of differentiation.</p>
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<p>We can derive the derivative of x(1/x) using proofs. To show this, we will use logarithmic differentiation 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 Logarithmic Differentiation </li>
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<ul><li>By Logarithmic Differentiation </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 Quotient Rule</li>
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<li>Using Quotient Rule</li>
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</ul><p>We will now demonstrate that the differentiation of x(1/x) results in x(1/x) * [(1 - ln(x))x²] using the above-mentioned methods:</p>
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</ul><p>We will now demonstrate that the differentiation of x(1/x) results in x(1/x) * [(1 - ln(x))x²] using the above-mentioned methods:</p>
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<h2>By Logarithmic Differentiation</h2>
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<h2>By Logarithmic Differentiation</h2>
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<p>The derivative of x(1/x) can be proved using logarithmic differentiation, which simplifies the differentiation process for functions raised to<a>variable</a><a>exponents</a>. To find the derivative of x(1/x), consider y = x(1/x). Taking the natural logarithm on both sides, ln(y) = (1/x) * ln(x) Differentiating both sides with respect to x, 1/y * dy/dx = (d/dx (ln(x))x) = (1/x²) - (ln(x)x²) dy/dx = y * [(1 - ln(x))x²] Since y = x(1/x), substitute back to get the derivative: dy/dx = x(1/x) * [(1 - ln(x))x²]</p>
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<p>The derivative of x(1/x) can be proved using logarithmic differentiation, which simplifies the differentiation process for functions raised to<a>variable</a><a>exponents</a>. To find the derivative of x(1/x), consider y = x(1/x). Taking the natural logarithm on both sides, ln(y) = (1/x) * ln(x) Differentiating both sides with respect to x, 1/y * dy/dx = (d/dx (ln(x))x) = (1/x²) - (ln(x)x²) dy/dx = y * [(1 - ln(x))x²] Since y = x(1/x), substitute back to get the derivative: dy/dx = x(1/x) * [(1 - ln(x))x²]</p>
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<h2>Using Chain Rule</h2>
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<h2>Using Chain Rule</h2>
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<p>To prove the differentiation of x(1/x) using the chain rule, We use the transformation: x(1/x) = e(ln(x)/x) Consider u(x) = ln(x)x, then y = eu By chain rule: dy/dx = eu * du/dx du/dx = (1/x²) - (ln(x)/x²) Thus, dy/dx = e(ln(x)/x) * [(1 - ln(x))/x²] Since e(ln(x)/x) = x(1/x), dy/dx = x(1/x) * [(1 - ln(x))/x²]</p>
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<p>To prove the differentiation of x(1/x) using the chain rule, We use the transformation: x(1/x) = e(ln(x)/x) Consider u(x) = ln(x)x, then y = eu By chain rule: dy/dx = eu * du/dx du/dx = (1/x²) - (ln(x)/x²) Thus, dy/dx = e(ln(x)/x) * [(1 - ln(x))/x²] Since e(ln(x)/x) = x(1/x), dy/dx = x(1/x) * [(1 - ln(x))/x²]</p>
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<h2>Using Quotient Rule</h2>
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<h2>Using Quotient Rule</h2>
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<p>We can rewrite the function as e^(ln(x)/x) for differentiation. Let u = ln(x) and v = x Therefore, y = u/v Using the<a>quotient</a>rule: d/dx (u/v) = (v * du/dx - u * dv/dx)/v² du/dx = 1/x and dv/dx = 1 Applying to the quotient rule: (1 * 1/x - ln(x) * 1)/x² = (1 - ln(x))/x² Thus, the derivative is: dy/dx = x(1/x) * [(1 - ln(x))/x²]</p>
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<p>We can rewrite the function as e^(ln(x)/x) for differentiation. Let u = ln(x) and v = x Therefore, y = u/v Using the<a>quotient</a>rule: d/dx (u/v) = (v * du/dx - u * dv/dx)/v² du/dx = 1/x and dv/dx = 1 Applying to the quotient rule: (1 * 1/x - ln(x) * 1)/x² = (1 - ln(x))/x² Thus, the derivative is: dy/dx = x(1/x) * [(1 - ln(x))/x²]</p>
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<h2>Higher-Order Derivatives of x^(1/x)</h2>
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<h2>Higher-Order Derivatives of x^(1/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. 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/x).</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. 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/x).</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/x), 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/x), 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 function x(1/x) is not defined, and thus its derivative is also undefined.</p>
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<p>When x is 0, the function x(1/x) is not defined, and thus its derivative is also undefined.</p>
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<p>For x = 1, the derivative of x(1/x) simplifies to 1, as x(1/x) = 1 and the derivative of a<a>constant</a>is 0.</p>
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<p>For x = 1, the derivative of x(1/x) simplifies to 1, as x(1/x) = 1 and the derivative of a<a>constant</a>is 0.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of x^(1/x)</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of x^(1/x)</h2>
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<p>Students frequently make mistakes when differentiating x(1/x). 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/x). 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/x)·x^2)</p>
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<p>Calculate the derivative of (x^(1/x)·x^2)</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Here, we have f(x) = x(1/x)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x(1/x) and v = x². Let’s differentiate each term, u′ = d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²] v′ = d/dx (x²) = 2x Substituting into the given equation, f'(x) = [x(1/x) * [(1 - ln(x))x²]] * x² + x(1/x) * 2x Let’s simplify terms to get the final answer, f'(x) = x(1/x) * (1 - ln(x)) + 2x(1/x+1) Thus, the derivative of the specified function is x(1/x) * (1 - ln(x)) + 2x(1/x+1).</p>
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<p>Here, we have f(x) = x(1/x)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x(1/x) and v = x². Let’s differentiate each term, u′ = d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²] v′ = d/dx (x²) = 2x Substituting into the given equation, f'(x) = [x(1/x) * [(1 - ln(x))x²]] * x² + x(1/x) * 2x Let’s simplify terms to get the final answer, f'(x) = x(1/x) * (1 - ln(x)) + 2x(1/x+1) Thus, the derivative of the specified function is x(1/x) * (1 - ln(x)) + 2x(1/x+1).</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.</p>
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<p>We find the derivative of the given function by dividing the function into two parts.</p>
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<p>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>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 is being inflated in the shape of a sphere, and its volume is given by V = x^(1/x) cubic meters, where x is the radius in meters. If the radius is 4 meters, find the rate at which the volume changes with respect to the radius.</p>
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<p>A balloon is being inflated in the shape of a sphere, and its volume is given by V = x^(1/x) cubic meters, where x is the radius in meters. If the radius is 4 meters, find the rate at which the volume changes with respect to the radius.</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 V = x(1/x) (volume of the balloon)...(1) Now, we will differentiate the equation (1) Take the derivative of x(1/x): dV/dx = x(1/x) * [(1 - ln(x))x²] Given x = 4 (substitute this into the derivative) dV/dx = 4(1/4) * [(1 - ln(4))/16] Calculating further, 4(1/4) = √2, ln(4) ≈ 1.386 dV/dx ≈ √2 * [(1 - 1.386)/16] dV/dx ≈ √2 * [-0.386/16] dV/dx ≈ -0.034 Hence, the rate at which the volume changes with respect to the radius at x = 4 meters is approximately -0.034 cubic meters per meter.</p>
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<p>We have V = x(1/x) (volume of the balloon)...(1) Now, we will differentiate the equation (1) Take the derivative of x(1/x): dV/dx = x(1/x) * [(1 - ln(x))x²] Given x = 4 (substitute this into the derivative) dV/dx = 4(1/4) * [(1 - ln(4))/16] Calculating further, 4(1/4) = √2, ln(4) ≈ 1.386 dV/dx ≈ √2 * [(1 - 1.386)/16] dV/dx ≈ √2 * [-0.386/16] dV/dx ≈ -0.034 Hence, the rate at which the volume changes with respect to the radius at x = 4 meters is approximately -0.034 cubic meters per meter.</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 volume at x = 4 meters by differentiating the volume function and substituting the given radius to calculate the rate at which the volume changes.</p>
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<p>We find the rate of change of volume at x = 4 meters by differentiating the volume function and substituting the given radius to calculate the rate at which the volume changes.</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/x).</p>
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<p>Derive the second derivative of the function y = x^(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>The first step is to find the first derivative, dy/dx = x(1/x) * [(1 - ln(x))x²]...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [x(1/x) * [(1 - ln(x))x²]] Using the product rule, d²y/dx² = [d/dx (x(1/x)) * (1 - ln(x))x²] + [x(1/x) * d/dx ((1 - ln(x))x²)] Calculating each part, d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²] d/dx ((1 - ln(x))x²) = [-1/x² - (1/x * -2/x³)] = [-1/x² + 2ln(x)x³] Substitute back, d²y/dx² = [x(1/x) * [(1 - ln(x))x²] * (1 - ln(x))x²] + [x(1/x) * [-1/x² + 2ln(x)x³]] Simplifying further will give the second derivative in more detailed form.</p>
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<p>The first step is to find the first derivative, dy/dx = x(1/x) * [(1 - ln(x))x²]...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [x(1/x) * [(1 - ln(x))x²]] Using the product rule, d²y/dx² = [d/dx (x(1/x)) * (1 - ln(x))x²] + [x(1/x) * d/dx ((1 - ln(x))x²)] Calculating each part, d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²] d/dx ((1 - ln(x))x²) = [-1/x² - (1/x * -2/x³)] = [-1/x² + 2ln(x)x³] Substitute back, d²y/dx² = [x(1/x) * [(1 - ln(x))x²] * (1 - ln(x))x²] + [x(1/x) * [-1/x² + 2ln(x)x³]] Simplifying further will give the second derivative in more detailed form.</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.</p>
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<p>We use the step-by-step process, where we start with the first derivative.</p>
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<p>Using the product rule, we differentiate the first derivative to find the second derivative.</p>
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<p>Using the product rule, we differentiate the first derivative to find the second derivative.</p>
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<p>We then substitute the identities and simplify the terms to find the final answer.</p>
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<p>We then substitute the identities and 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 (x^(1/x+1)) = x^(1/x+1) * [(1 - ln(x+1))/(x+1)²].</p>
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<p>Prove: d/dx (x^(1/x+1)) = x^(1/x+1) * [(1 - ln(x+1))/(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>Let’s start using the chain rule: Consider y = x(1/x+1) Rewrite y = e((ln(x))/(x+1)) Differentiate using the chain rule: dy/dx = e((ln(x))/(x+1)) * d/dx ((ln(x))/(x+1)) The derivative of (ln(x))/(x+1) using quotient rule: d/dx ((ln(x))/(x+1)) = [(x+1)(1/x) - ln(x)(1)]/(x+1)² = [1 - ln(x)]/(x+1)² Substituting back, dy/dx = x(1/x+1) * [(1 - ln(x))/(x+1)²] Hence proved.</p>
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<p>Let’s start using the chain rule: Consider y = x(1/x+1) Rewrite y = e((ln(x))/(x+1)) Differentiate using the chain rule: dy/dx = e((ln(x))/(x+1)) * d/dx ((ln(x))/(x+1)) The derivative of (ln(x))/(x+1) using quotient rule: d/dx ((ln(x))/(x+1)) = [(x+1)(1/x) - ln(x)(1)]/(x+1)² = [1 - ln(x)]/(x+1)² Substituting back, dy/dx = x(1/x+1) * [(1 - ln(x))/(x+1)²] 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.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation.</p>
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<p>Then, we replace the expression with its derivative.</p>
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<p>Then, we replace the expression with its derivative.</p>
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<p>As a final step, we substitute y = x(1/x+1) to derive the equation.</p>
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<p>As a final step, we substitute y = x(1/x+1) 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^(1/x)/x)</p>
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<p>Solve: d/dx (x^(1/x)/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 (x(1/x)/x) = (d/dx (x(1/x)) * x - x(1/x) * d/dx(x))x² We will substitute d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²] and d/dx (x) = 1 = [x(1/x) * [(1 - ln(x))x²] * x - x(1/x)]/x² = [x(1/x) * (1 - ln(x)) - x(1/x)]/x³ = x(1/x) * [(1 - ln(x)) - 1]/x³ Therefore, d/dx (x(1/x)/x) = x(1/x) * [-ln(x)]/x³</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (x(1/x)/x) = (d/dx (x(1/x)) * x - x(1/x) * d/dx(x))x² We will substitute d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²] and d/dx (x) = 1 = [x(1/x) * [(1 - ln(x))x²] * x - x(1/x)]/x² = [x(1/x) * (1 - ln(x)) - x(1/x)]/x³ = x(1/x) * [(1 - ln(x)) - 1]/x³ Therefore, d/dx (x(1/x)/x) = x(1/x) * [-ln(x)]/x³</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.</p>
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<p>In this process, we differentiate the given function using the quotient rule.</p>
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<p>As a final step, we simplify the equation to obtain the final result.</p>
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<p>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/x)</h2>
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<h2>FAQs on the Derivative of x^(1/x)</h2>
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<h3>1.Find the derivative of x^(1/x).</h3>
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<h3>1.Find the derivative of x^(1/x).</h3>
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<p>Using logarithmic differentiation for x^(1/x), d/dx (x^(1/x)) = x^(1/x) * [(1 - ln(x))/x²]</p>
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<p>Using logarithmic differentiation for x^(1/x), d/dx (x^(1/x)) = x^(1/x) * [(1 - ln(x))/x²]</p>
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<h3>2.Can we use the derivative of x^(1/x) in real life?</h3>
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<h3>2.Can we use the derivative of x^(1/x) in real life?</h3>
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<p>Yes, we can use the derivative of x^(1/x) in real life to analyze rates of change in various fields such as mathematics, physics, and economics.</p>
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<p>Yes, we can use the derivative of x^(1/x) in real life to analyze rates of change in various fields such as mathematics, physics, and economics.</p>
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<h3>3.Is it possible to take the derivative of x^(1/x) at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of x^(1/x) at the point where x = 0?</h3>
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<p>No, x = 0 is a point where x^(1/x) is undefined, so it is impossible to take the derivative at these points (since the function does not exist there).</p>
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<p>No, x = 0 is a point where x^(1/x) is undefined, so it is impossible to take the derivative at these points (since the function does not exist there).</p>
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<h3>4.What rule is used to differentiate x^(1/x)/x?</h3>
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<h3>4.What rule is used to differentiate x^(1/x)/x?</h3>
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<p>We use the quotient rule to differentiate x^(1/x)/x, d/dx (x^(1/x)/x) = [x^(1/x) * [(1 - ln(x))/x²] * x - x^(1/x)]/x².</p>
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<p>We use the quotient rule to differentiate x^(1/x)/x, d/dx (x^(1/x)/x) = [x^(1/x) * [(1 - ln(x))/x²] * x - x^(1/x)]/x².</p>
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<h3>5.Are the derivatives of x^(1/x) and x^(-1/x) the same?</h3>
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<h3>5.Are the derivatives of x^(1/x) and x^(-1/x) the same?</h3>
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<p>No, they are different. The derivative of x^(1/x) is x^(1/x) * [(1 - ln(x))/x²], while the derivative of x^(-1/x) involves different expressions due to the<a>negative exponent</a>.</p>
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<p>No, they are different. The derivative of x^(1/x) is x^(1/x) * [(1 - ln(x))/x²], while the derivative of x^(-1/x) involves different expressions due to the<a>negative exponent</a>.</p>
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<h3>6.Can we find the derivative of the x^(1/x) formula?</h3>
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<h3>6.Can we find the derivative of the x^(1/x) formula?</h3>
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<p>To find it, consider y = x^(1/x). We use logarithmic differentiation: ln(y) = ln(x)/x Differentiating, dy/dx = x^(1/x) * [(1 - ln(x))/x²].</p>
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<p>To find it, consider y = x^(1/x). We use logarithmic differentiation: ln(y) = ln(x)/x Differentiating, dy/dx = x^(1/x) * [(1 - ln(x))/x²].</p>
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<h2>Important Glossaries for the Derivative of x^(1/x)</h2>
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<h2>Important Glossaries for the Derivative of x^(1/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>Logarithmic Differentiation:</strong>A method used to differentiate functions with variable exponents by taking the natural log of both sides.</li>
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</ul><ul><li><strong>Logarithmic Differentiation:</strong>A method used to differentiate functions with variable exponents by taking the natural log of both sides.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Power Function:</strong>A function where the variable is raised to a power, such as x(1/x).</li>
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</ul><ul><li><strong>Power Function:</strong>A function where the variable is raised to a power, such as x(1/x).</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating the ratio of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating the ratio of two functions.</li>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<p>▶</p>
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<p>▶</p>
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<h2>Jaskaran Singh Saluja</h2>
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<h2>Jaskaran Singh Saluja</h2>
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<h3>About the Author</h3>
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<h3>About the Author</h3>
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<p>Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.</p>
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<p>Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.</p>
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<h3>Fun Fact</h3>
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<h3>Fun Fact</h3>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>