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战巡
发表于 2015-11-2 13:18
回复 1# hjnk900
这不基础题么?
易求
\[x=\frac{1}{2}\ln(\frac{u+v}{1+e^{2w}})+w\]
\[\frac{\partial x}{\partial u}=\frac{1}{2(u+v)}=\frac{1}{2(e^{2x}+e^{2y})}\]
\[\frac{\partial x}{\partial v}=\frac{1}{2(u+v)}=\frac{1}{2(e^{2x}+e^{2y})}\]
\[\frac{\partial x}{\partial w}=\frac{1}{1+e^{2w}}=\frac{1}{1+e^{2(x-y)}}\] |
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