http://dhi.org.mx/wiki/api.php?action=feedcontributions&feedformat=atom&user=ReynaNesbit5549 2026-05-02T02:48:41Z Contribuciones del usuario MediaWiki 1.19.20 http://dhi.org.mx/wiki/index.php?title=Trang_Websex_Hang_Dau 2025-06-05T13:12:13Z

<p>ReynaNesbit5549: </p> <hr /> <div>[https://88888clb.com sex người lớn] siêu hấp dẫn, các em chân dài khắp mọi nơi</div>

ReynaNesbit5549 http://dhi.org.mx/wiki/index.php?title=Usuario:ReynaNesbit5549 2025-06-05T13:12:09Z

<p>ReynaNesbit5549: Página creada con «I don't believe that the answer is ln(x)x^(ln(x)-2), since the power rule doesn't apply when you have the variable in the exponent. Do the following instead:y x^ln(x)&lt;br&gt;&lt;b...»</p> <hr /> <div>I don't believe that the answer is ln(x)x^(ln(x)-2), since the power rule doesn't apply when you have the variable in the exponent. Do the following instead:y x^ln(x)&lt;br&gt;&lt;br&gt;Taking the natural log of both sides:ln(y)ln(x) * ln(x)ln(y) ln(x)^2&lt;br&gt;&lt;br&gt;Take the derivative of both sides, using the chain rule:1/y * y' 2 ln(x) / xy' 2 ln(x)/ x * y&lt;br&gt;&lt;br&gt;Finally, substitute in the first equation, y x^ln(x):y' 2 ln(x) / x * x^ln(x)y'2 ln(x) * x ^ (ln(x) - 1)&lt;br&gt;&lt;br&gt;Sorry if everything is formatted really badly, this is my first post on answers.com.&lt;br&gt;&lt;br&gt;my web page [https://88888clb.com sex người lớn]</div>

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