设z=sinx+F(siny-sinx),其中F为可微函数,求证:(偏z⼀偏x)cosy+(偏z⼀偏y)cosx=cosxcosy

答案是其次的,主要是过程,越详细越好!
2025-06-20 18:24:39
推荐回答(1个)
回答1:

dz/dx=cosx+(dF(siny-sinx)/dx)*(-cosx)
dz/dy=(dF(siny-sinx)/dy)*(cosy)
(dz/dx)cosy+(dz/dy)cosx=[cosx+(dF(siny-sinx)/dx)*(-cosx)]cosy+(dF(siny-sinx)/dy)*(cosy)cosx
=cosxcosy-(dF(siny-sinx)/dx)*(cosxcosy)+(dF(siny-sinx)/dx)*(cosxcosy)
=cosxcosy