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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>