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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]

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<p>To solve the problem, we need to find <span class="mjx-chtml MJXc-display" style="text-align: center;"><span class="mjx-math"><span class="mjx-mrow"><span class="mjx-msup"><span class="mjx-base" style="margin-right: -0.006em;"><span class="mjx-mi"><span class="mjx-char MJXc-TeX-math-I" style="padding-top: 0.225em; padding-bottom: 0.519em; padding-right: 0.006em;">y</span></span></span><span class="mjx-sup" style="font-size: 70.7%; vertical-align: 0.584em; padding-left: 0.082em; padding-right: 0.071em;"><span class="mjx-mo" style=""><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.298em; padding-bottom: 0.298em;">′′</span></span></span></span><span class="mjx-mo"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.446em; padding-bottom: 0.593em;">(</span></span><span class="mjx-mn"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.372em; padding-bottom: 0.372em;">2</span></span><span class="mjx-mo"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.446em; padding-bottom: 0.593em;">)</span></span><span class="mjx-mo MJXc-space2"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.298em; padding-bottom: 0.446em;">−</span></span><span class="mjx-mn MJXc-space2"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.372em; padding-bottom: 0.372em;">2</span></span><span class="mjx-msup"><span class="mjx-base" style="margin-right: -0.006em;"><span class="mjx-mi"><span class="mjx-char MJXc-TeX-math-I" style="padding-top: 0.225em; padding-bottom: 0.519em; padding-right: 0.006em;">y</span></span></span><span class="mjx-sup" style="font-size: 70.7%; vertical-align: 0.584em; padding-left: 0.082em; padding-right: 0.071em;"><span class="mjx-mo" style=""><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.298em; padding-bottom: 0.298em;">′</span></span></span></span><span class="mjx-mo"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.446em; padding-bottom: 0.593em;">(</span></span><span class="mjx-mn"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.372em; padding-bottom: 0.372em;">2</span></span><span class="mjx-mo"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.446em; padding-bottom: 0.593em;">)</span></span></span></span></span> for the function <span class="mjx-chtml MJXc-display" style="text-align: center;"><span class="mjx-math"><span class="mjx-mrow"><span class="mjx-mi"><span class="mjx-char MJXc-TeX-math-I" style="padding-top: 0.225em; padding-bottom: 0.519em; padding-right: 0.006em;">y</span></span><span class="mjx-mo"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.446em; padding-bottom: 0.593em;">(</span></span><span class="mjx-mi"><span class="mjx-char MJXc-TeX-math-I" style="padding-top: 0.225em; padding-bottom: 0.298em;">x</span></span><span class="mjx-mo"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.446em; padding-bottom: 0.593em;">)</span></span><span class="mjx-mo MJXc-space3"><span class="mjx-char MJXc-TeX-main-R" style="padding-top: 0.077em; padding-bottom: 0.298em;">=</span></span><span class="mjx-msubsup MJXc-space3"><span class="mjx-base"><span class="mjx-mi"><span class="mjx-char MJXc-TeX-math-I" style="padding-top: 0.225em; padding-bottom: 0.298em;">x</span></span></span><span class="mjx-sup" style="font-size: 70.7%; vertical-align: 0.584em; padding-left: 0px; padding-right: 0.071em;"><span class="mjx-mi" style=""><span class="mjx-char MJXc-TeX-math-I" style="padding-top: 0.225em; padding-bottom: 0.298em;">x</span></span></span></span></span></span></span>.</p><p><strong>Step 1: Find \( y'(x) \)</strong></p><p>We start by differentiating \( y(x) = x^x \). To do this, we can use logarithmic differentiation.</p><p>1. Take the natural logarithm of both sides: \( \ln y = \ln(x^x) = x \ln x \)</p><p>2. Differentiate both sides with respect to \( x \): \( \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \)</p><p>3. Multiply through by \( y \): \( y' = y(\ln x + 1) = x^x(\ln x + 1) \)</p><p><strong>Step 2: Find \( y'(2) \)</strong></p><p>Now we substitute \( x = 2 \): \( y'(2) = 2^2(\ln 2 + 1) = 4(\ln 2 + 1) \)</p><p><strong>Step 3: Find \( y''(x) \)</strong></p><p>Next, we differentiate \( y'(x) \) to find \( y''(x) \). We will use the product rule: \( y' = x^x(\ln x + 1) \)</p><p>Using the product rule: \( y'' = \frac{d}{dx}(x^x) \cdot (\ln x + 1) + x^x \cdot \frac{d}{dx}(\ln x + 1) \)</p><p>We already found \( \frac{d}{dx}(x^x) = x^x(\ln x + 1) \), and: \( \frac{d}{dx}(\ln x + 1) = \frac{1}{x} \)</p><p>Thus: \( y'' = x^x(\ln x + 1)(\ln x + 1) + x^x \cdot \frac{1}{x} \) \( = x^x(\ln x + 1)^2 + x^{x-1} \)</p><p><strong>Step 4: Find \( y''(2) \)</strong></p><p>Now we substitute \( x = 2 \): \( y''(2) = 2^2(\ln 2 + 1)^2 + 2^{2-1} = 4(\ln 2 + 1)^2 + 2 \)</p><p><strong>Step 5: Calculate \( y''(2) - 2y'(2) \)</strong></p><p>Now we can calculate: \( y''(2) - 2y'(2) = (4(\ln 2 + 1)^2 + 2) - 2(4(\ln 2 + 1)) \) \( = 4(\ln 2 + 1)^2 + 2 - 8(\ln 2 + 1) \)</p><p><strong>Step 6: Simplify the expression</strong></p><p>Let’s simplify: \( = 4(\ln 2 + 1)^2 - 8(\ln 2 + 1) + 2 \)</p><p>This is a quadratic in terms of \( \ln 2 + 1 \): Let \( u = \ln 2 + 1 \): \( = 4u^2 - 8u + 2 \)</p><p><strong>Step 7: Factor or use the quadratic formula</strong></p><p>We can use the quadratic formula: \( = 4(u^2 - 2u + \frac{1}{2}) = 4\left((u - 1)^2 - \frac{1}{2}\right) \)</p><p><strong>Final Result</strong></p><p>Thus, the final answer is: \( y''(2) - 2y'(2) = 4\left((\ln 2 + 1 - 1)^2 - \frac{1}{2}\right) = 4\left((\ln 2)^2 - \frac{1}{2}\right) \)</p>

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Looking for baby boy haircuts? We have both traditional little boy styles and copies of adults’ cuts in our gallery. The level of cuteness of these baby haircuts has exceeded the legal limit…

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]

<p>Tìm x, biết:| 2022 - x | + | 2023 - x l + l 2024 - x | = 2 giúp mình vs</p>