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If y(x) = x^x , x gt 0 then y"(2) – 2y'(2) is equal to
<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>
Pornography prayer points with scriptures - Prayer Points for Today
Breaking free from pornography is a process that requires prayer, accountability, and practical steps like avoiding triggers and seeking counseling if needed. Remember, God’s grace is sufficient, and His power is made perfect in our weakness (2 Corinthians 12:9). Below are 10 pornography prayer points with scriptures to help break free from pornography, each supported by relevant Bible verses.