为什么不是y乘f'u直接对x求导，而是第三幅图那样复合的。没搞懂，老师说f'(u,v)依然是u,v的函数为什么？ 复变函数求导 f(z)=u+iv,u=u(x,y),v=v(...
\u8bbez=y/f(x²-y²),\u5176\u4e2df(u)\u4e3a\u53ef\u5bfc\u51fd\u6570\uff0c\u6c42z\u5bf9x\u548cy\u7684\u504f\u5bfc\u6570\u89e3\uff1a
\u5206\u6790\uff0c\u7ba1\u5b83\u51e0\u5c42\u590d\u5408\uff0c\u8fd9\u79cd\u9898\uff0c\u8c01\u8fd8\u8ba1\u8f83\u590d\u5408\u4e0d\u590d\u5408\uff1f\u504f\u5bfc\u9898\u90fd\u662f\u5148\u5224\u65ad\u81ea\u53d8\u91cf\uff0c\u548c\u56e0\u53d8\u91cf\uff0c\u81f3\u4e8e\u590d\u5408\u7684\u5c42\u6570\uff0c\u4e0d\u7528\u5173\u5fc3\uff01\u603b\u4e4b\uff0c\u5c31\u662f\u94fe\u5f0f\u6cd5\u5219\u5c31\u5bf9\u4e86\uff01
∂z/∂x
=-y\u00b7{∂[f(x²-y²)]/∂x}/f²(x²-y²)
=-y\u00b7{f'(x²-y²)\u00b7[∂(x²-y²)/∂x]}/f²(x²-y²)
=-y\u00b7[f'(x²-y²)\u00b72x]/f²(x²-y²)
=-2xyf'(x²-y²)/f²(x²-y²)
\u94fe\u5f0f\u6cd5\u5219\uff1a
∂z/∂x=(∂z/∂u)\u00b7(∂u/∂v)\u00b7(∂v/∂t)\u00b7(∂t/∂p).........\u00b7(∂m/∂x)
\u5176\u4e2d\uff1a∂u/∂v,∂v/∂t,∂t/∂p,.........∂m/∂x\u5747\u5b58\u5728\u5168\u5fae\u5206
\u4e0a\u8ff0\u5f88\u7b80\u5355\uff0c\u53ef\u7528\u504f\u5bfc\u5b9a\u4e49\u8bc1\u660e\uff01
\u4e0d\u662f\u6240\u6709\u7684\u590d\u53d8\u51fd\u6570\u90fd\u662f\u89e3\u6790\u7684\uff0c\u5982\u679c\u590d\u53d8\u51fd\u6570\u89e3\u6790\uff0c\u90a3\u4e48\u5b83\u5c31\u6ee1\u8db3C-R\u65b9\u7a0b\uff0c\u5373Ux=Vy,Vx=-Uy\uff0c\u6240\u4ee5\u5bf9x\u548c\u5bf9y\u7684\u504f\u5bfc\u6570\u53ef\u4ee5\u76f8\u4e92\u8868\u793a\uff0c\u4e3a\u4e86\u65b9\u4fbf\u4e00\u822c\u5c31\u7528\u5bf9x\u7684\u504f\u5bfc\u6570\u6765\u8868\u793a\u4e86\u800c\u5df2\uff0c\u5176\u5b9e\u7528y\u4e5f\u662f\u53ef\u4ee5\u7684\u3002\u671b\u91c7\u7eb3\uff0c\u8c22\u8c22\uff01
因为一阶偏导还是含有未知函数，再进行二阶求导是还是将其看作未知数来求解。。绛旓細浣犲枩娆㈠摢绉嶅氨鐢ㄥ摢绉嶃備互鍚庤繕浼氶亣鍒板涓嚜鍙橀噺鐨勬儏鍐碉紝濡倄銆y鏄嚜鍙橀噺锛岄偅灏辩敤z鏉ヨ〃绀哄嚱鏁帮紝濡倆=sinx+siny锛屽綋鐒朵篃鍙互鐢f(x,y)=sinx+siny鏉ヨ〃绀恒備功涓綔鑰呯敤鍓嶈咃紝璇存槑浠栧枩娆㈢敤鍓嶈咃紝浣嗕篃鏈変笉灏戜綔鑰呯敤鍚庤咃紝鎴栬呬袱鑰呴兘鐢紝杩欏彧鏄範鎯屽凡锛屾病鏈夊お澶氱殑璁茬┒銆
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绛旓細鍥犱负鏄簩閲
