方程式:y=cos{atan[x/(1-x)^(1/2) ]} 求偏导:dy/dx
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方程式:y=cos{atan[x/(1-x)^(1/2) ]} 求偏导:dy/dx
![方程式:y=cos{atan[x/(1-x)^(1/2) ]} 求偏导:dy/dx](/uploads/image/z/16619457-57-7.jpg?t=%E6%96%B9%E7%A8%8B%E5%BC%8F%EF%BC%9Ay%3Dcos%7Batan%5Bx%2F%281-x%29%5E%281%2F2%29+%5D%7D+%E6%B1%82%E5%81%8F%E5%AF%BC%EF%BC%9Ady%2Fdx)
dy/dx
=-sin{atan[x/(1-x)^(1/2) ]}*1/{1+[x/(1-x)^(1/2) ]^2}*[(1-x)^(1/2)+x/(1-x)^(1/2)]/(1-x)
再问: 是否应该:=-sin{atan[x/(1-x)^(1/2) ]}*1/{1+[x/(1-x)^(1/2) ]^2}*[(1-x)^(1/2)+ “1/2” x/(1-x)^(1/2)]/(1-x) ; 似乎缺少1/2系数,确认下!
再答: 是的,漏写了
=-sin{atan[x/(1-x)^(1/2) ]}*1/{1+[x/(1-x)^(1/2) ]^2}*[(1-x)^(1/2)+x/(1-x)^(1/2)]/(1-x)
再问: 是否应该:=-sin{atan[x/(1-x)^(1/2) ]}*1/{1+[x/(1-x)^(1/2) ]^2}*[(1-x)^(1/2)+ “1/2” x/(1-x)^(1/2)]/(1-x) ; 似乎缺少1/2系数,确认下!
再答: 是的,漏写了