Найдите производную:[tex] \displaystyle y= \frac{3^x}{2^x+5^x}[/tex]. Created by MrsVaderr. algebra-ru

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$y'=(\frac{3^{x}}{2^{x}+5^{x}})'=\frac{(3^{x})'*(2^{x}+5^{x})-3^{x}*(2^{x}+5^{x})'}{(2^{x}+5^{x})^{2}}=\frac{3^{x}*ln3*(2^{x}+5^{x})-3^{x}*(2^{x}*ln2+5^{x}*ln5)}{(2^{x}+5^{x})^{2}}=\frac{3^{x}(2^{x}*ln3+5^{x}*ln3-2^{x}*ln2-5^{x}*ln5)}{(2^{x}+5^{x})^{2}} =\frac{3^{x}(2^{x}*ln1,5+5^{x}*ln0,6)}{(2^{x}+5^{x})^{2}}$ $=\frac{6^{x}*ln1,5+15^{x}*ln0,6}{(2^{x}+5^{x})^{2}}}$

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Y' =
$\frac{ {3}^{x} ln3 \times ( {2}^{x} + {5}^{x} ) - {3}^{x} ( {2}^{x} ln2 + {5}^{x}ln5) }{ {( {2}^{x} + {5}^{x} )}^{2} } = \frac{ {3}^{x} {2}^{x}ln3 + {3}^{x} {5}^{x} ln3 - {3}^{x} {2}^{x} ln2 - {3}^{x} {5}^{x}ln5 }{ {( {2}^{x} + {5}^{x} )}^{2} } = \frac{ {3}^{x} {2}^{x}(ln3 - ln2) + {3}^{x} {5}^{x}(ln3 - ln5) }{ {( {2}^{x} + {5}^{x} )}^{2} } = \frac{ {6}^{x}ln \frac{3}{2} + {15}^{x} ln \frac{3}{5} }{ {( {2}^{x} + {5}^{x} )}^{2} }$

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