sinxy-ln(x+y)=0两边对X求导 cos(xy)(y+xy')-(1+y‘)/(x+y)=0 其中(1+y‘)
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sinxy-ln(x+y)=0两边对X求导 cos(xy)(y+xy')-(1+y‘)/(x+y)=0 其中(1+y‘)/(x+y)怎么得来的?
![sinxy-ln(x+y)=0两边对X求导 cos(xy)(y+xy')-(1+y‘)/(x+y)=0 其中(1+y‘)](/uploads/image/z/17982995-59-5.jpg?t=sinxy-ln%28x%2By%29%3D0%E4%B8%A4%E8%BE%B9%E5%AF%B9X%E6%B1%82%E5%AF%BC+cos%28xy%29%28y%2Bxy%27%29-%281%2By%E2%80%98%29%2F%28x%2By%29%3D0+%E5%85%B6%E4%B8%AD%281%2By%E2%80%98%29)
ln(x+y)求导=1/(x+y)*(x+y)'
=1/(x+y)*(1+y')
=(1+y')/(x+y)
=1/(x+y)*(1+y')
=(1+y')/(x+y)
讲教一下师傅隐函数求导sinxy-ln(x+y)=0
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