求隐函数e^(xy)+ln(y/(x+1))=0的二阶导数
<p>问题：求隐函数e^(xy)+ln(y/(x+1))=0的二阶导数<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">胡朝晖的回答：<div class="content-b">网友采纳 e^(xy)+ln(y/(x+1))=e^(xy)+lny-ln(x+1))=0两边求导,记住y是x的函数,它在这里应看作中间变量,用复合函数求导法则∴e^(xy)*(y+xy')+y'/y-1/(x+1)=0⑴y'=-ye^(xy)+1/(x+1),∴y'=[-ye^(xy)+1/(x+1)/[1/y+...
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