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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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If you’re scrolling through Instagram or Pinterest, you’ve probably seen the girl hidden face blurry aesthetic mirror selfie trend. It’s mysterious, artsy, and effortlessly cool. But how do you get that perfect shot? Let’s break it down step by step.

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