x * e^y+siny=0
e^y + x * e^y * y' + cosy * y' = 0
=> y' = - e^y / [ x e^y + cosy ]你好!我數(shù)學(xué)太爛。。能不能補(bǔ)充一下完整的答案。。。x * e^y+siny=0兩邊同時對x求導(dǎo),注意y 是x 的函數(shù),(e^y) ' = (e^y * y', (siny)'=cosy * y'e^y + x * e^y * y' + cosy * y' = 0即( x e^y + cosy ) * y' = - e^ydy/dx=y' ……
x*e^y+siny=0 求dy/dx
x*e^y+siny=0 求dy/dx
數(shù)學(xué)人氣:390 ℃時間:2020-04-28 03:26:38
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