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10 Korean Swear Words You Should Know (With Examples) 한국의 욕 Korean has its own collection of swear and curse words, be it a local slang or a specific term, just like any other language. Disclaimer: Before we korean swear words

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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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