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Lịch sử đối đầu MU vs Crystal Palace: Tỷ số đậm

MU vs Crystal Palace không phải là hai đội bóng có sự kình địch lớn tại Premier League, nhưng mỗi lần chạm trán giữa họ đều mang đến những diễn biến đáng chú ý.

05:37 04/07/2025

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05:40 04/07/2025

If y(x) = x^x , x gt 0 then y"(2) – 2y'(2) is equal to

To solve the problem, we need to find y′′(2)−2y′(2) for the function y(x)=xx.Step 1: Find \( y'(x) \)We start by differentiating \( y(x) = x^x \). To do this, we can use logarithmic differentiation.1. Take the natural logarithm of both sides: \( \ln y = \ln(x^x) = x \ln x \)2. Differentiate both sides with respect to \( x \): \( \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \)3. Multiply through by \( y \): \( y' = y(\ln x + 1) = x^x(\ln x + 1) \)Step 2: Find \( y'(2) \)Now we substitute \( x = 2 \): \( y'(2) = 2^2(\ln 2 + 1) = 4(\ln 2 + 1) \)Step 3: Find \( y''(x) \)Next, we differentiate \( y'(x) \) to find \( y''(x) \). We will use the product rule: \( y' = x^x(\ln x + 1) \)Using the product rule: \( y'' = \frac{d}{dx}(x^x) \cdot (\ln x + 1) + x^x \cdot \frac{d}{dx}(\ln x + 1) \)We already found \( \frac{d}{dx}(x^x) = x^x(\ln x + 1) \), and: \( \frac{d}{dx}(\ln x + 1) = \frac{1}{x} \)Thus: \( y'' = x^x(\ln x + 1)(\ln x + 1) + x^x \cdot \frac{1}{x} \) \( = x^x(\ln x + 1)^2 + x^{x-1} \)Step 4: Find \( y''(2) \)Now we substitute \( x = 2 \): \( y''(2) = 2^2(\ln 2 + 1)^2 + 2^{2-1} = 4(\ln 2 + 1)^2 + 2 \)Step 5: Calculate \( y''(2) - 2y'(2) \)Now we can calculate: \( y''(2) - 2y'(2) = (4(\ln 2 + 1)^2 + 2) - 2(4(\ln 2 + 1)) \) \( = 4(\ln 2 + 1)^2 + 2 - 8(\ln 2 + 1) \)Step 6: Simplify the expressionLet’s simplify: \( = 4(\ln 2 + 1)^2 - 8(\ln 2 + 1) + 2 \)This is a quadratic in terms of \( \ln 2 + 1 \): Let \( u = \ln 2 + 1 \): \( = 4u^2 - 8u + 2 \)Step 7: Factor or use the quadratic formulaWe can use the quadratic formula: \( = 4(u^2 - 2u + \frac{1}{2}) = 4\left((u - 1)^2 - \frac{1}{2}\right) \)Final ResultThus, the final answer is: \( y''(2) - 2y'(2) = 4\left((\ln 2 + 1 - 1)^2 - \frac{1}{2}\right) = 4\left((\ln 2)^2 - \frac{1}{2}\right) \)

05:41 04/07/2025

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05:42 04/07/2025

Sexy, Sexy, Sexy by Julio de la Rosa on Apple Music

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05:44 04/07/2025

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05:46 04/07/2025

Julio de la Rosa - Sexy Sexy Sexy Lyrics

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05:50 04/07/2025

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05:52 04/07/2025

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05:53 04/07/2025

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06:00 04/07/2025

Jake Paul vs. Mike Tyson

Jake Paul vs. Mike Tyson was a heavyweight professional boxing match between YouTuber and professional boxer Jake Paul and former undisputed heavyweight world champion Mike Tyson. The bout took place on November 15, 2024, at the AT&T Stadium in Arlington, Texas, and was streamed globally on Netflix, with 65 million people watching the event concurrently[4][5] making it the most-streamed sporting event ever at the time.[6] Paul defeated Tyson via unanimous decision.[3]

06:01 04/07/2025

If y(x) = x^x , x gt 0 then y"(2) – 2y'(2) is equal to

To solve the problem, we need to find y′′(2)−2y′(2) for the function y(x)=xx.Step 1: Find \( y'(x) \)We start by differentiating \( y(x) = x^x \). To do this, we can use logarithmic differentiation.1. Take the natural logarithm of both sides: \( \ln y = \ln(x^x) = x \ln x \)2. Differentiate both sides with respect to \( x \): \( \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \)3. Multiply through by \( y \): \( y' = y(\ln x + 1) = x^x(\ln x + 1) \)Step 2: Find \( y'(2) \)Now we substitute \( x = 2 \): \( y'(2) = 2^2(\ln 2 + 1) = 4(\ln 2 + 1) \)Step 3: Find \( y''(x) \)Next, we differentiate \( y'(x) \) to find \( y''(x) \). We will use the product rule: \( y' = x^x(\ln x + 1) \)Using the product rule: \( y'' = \frac{d}{dx}(x^x) \cdot (\ln x + 1) + x^x \cdot \frac{d}{dx}(\ln x + 1) \)We already found \( \frac{d}{dx}(x^x) = x^x(\ln x + 1) \), and: \( \frac{d}{dx}(\ln x + 1) = \frac{1}{x} \)Thus: \( y'' = x^x(\ln x + 1)(\ln x + 1) + x^x \cdot \frac{1}{x} \) \( = x^x(\ln x + 1)^2 + x^{x-1} \)Step 4: Find \( y''(2) \)Now we substitute \( x = 2 \): \( y''(2) = 2^2(\ln 2 + 1)^2 + 2^{2-1} = 4(\ln 2 + 1)^2 + 2 \)Step 5: Calculate \( y''(2) - 2y'(2) \)Now we can calculate: \( y''(2) - 2y'(2) = (4(\ln 2 + 1)^2 + 2) - 2(4(\ln 2 + 1)) \) \( = 4(\ln 2 + 1)^2 + 2 - 8(\ln 2 + 1) \)Step 6: Simplify the expressionLet’s simplify: \( = 4(\ln 2 + 1)^2 - 8(\ln 2 + 1) + 2 \)This is a quadratic in terms of \( \ln 2 + 1 \): Let \( u = \ln 2 + 1 \): \( = 4u^2 - 8u + 2 \)Step 7: Factor or use the quadratic formulaWe can use the quadratic formula: \( = 4(u^2 - 2u + \frac{1}{2}) = 4\left((u - 1)^2 - \frac{1}{2}\right) \)Final ResultThus, the final answer is: \( y''(2) - 2y'(2) = 4\left((\ln 2 + 1 - 1)^2 - \frac{1}{2}\right) = 4\left((\ln 2)^2 - \frac{1}{2}\right) \)

06:07 04/07/2025

CONFIRMED lineups: Barcelona vs Real Madrid, 2025 El Clasico

Lịch sử đối đầu MU vs Crystal Palace: Tỷ số đậm

🎰 Slot Gacor Microstar88.MPO: Guide to Winning Big in 2025

If y(x) = x^x , x gt 0 then y"(2) – 2y'(2) is equal to

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Sexy, Sexy, Sexy by Julio de la Rosa on Apple Music

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Jake Paul vs. Mike Tyson

If y(x) = x^x , x gt 0 then y"(2) – 2y'(2) is equal to

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