[cos(xy)]的平方，如何分别对x和y求导？ [cos(xy)]的平方，如何分别对x和y求导？
[cos(xy)]\u7684\u5e73\u65b9,\u5982\u4f55\u5206\u522b\u5bf9x\u548cy\u6c42\u5bfc?cos²(xy),\u53ef\u4ee5\u770b\u505a\u662f\u590d\u5408\u51fd\u6570\uff1a
f(g)=g²,g(\u03c6)=cos(\u03c6),\u03c6(x)=xy
f'(x)=[f'(g)][g'(\u03c6)][\u03c6'(x)]
\u56e0\u6b64\uff1a
d[cos²(xy)]/dx=[2cos(xy)][cos'(xy)][(xy)']
=[2cos(xy)][-sin(xy)](y)
=-2ycos(xy)sin(xy)
=-ysin(2xy)
\u540c\u6837\u7684,\u6709\uff1a
d[cos²(xy)]/dy=-xsin(2xy)
\u89e3\uff1a
cos²(xy)\uff0c\u53ef\u4ee5\u770b\u505a\u662f\u590d\u5408\u51fd\u6570\uff1a
f(g)=g²\uff0cg(\u03c6)=cos(\u03c6)\uff0c\u03c6(x)=xy
f'(x)=[f'(g)][g'(\u03c6)][\u03c6'(x)]
\u56e0\u6b64\uff1a
d[cos²(xy)]/dx=[2cos(xy)][cos'(xy)][(xy)']
=[2cos(xy)][-sin(xy)](y)
=-2ycos(xy)sin(xy)
=-ysin(2xy)
\u540c\u6837\u7684\uff0c\u6709\uff1a
d[cos²(xy)]/dy=-xsin(2xy)
[f(g(x))]'=f'(g(x))g'(x)
此处
1)f(z)=z^2,g(x)=cos(xy)
f'(z)=2z
所以
f'(g(x))g'(x)
=[2*cos(xy)]*[-sin(xy)*(xy)']
右边括号再次使用了链式(f(z)=sin(z),g(x)=xy)
=[2*cos(xy)]*[-ysin(xy)]
2)同理可得对y的偏导数为[2*cos(xy)]*[-xsin(xy)]
解：
cos²(xy)，可以看做是复合函数：
f(g)=g²，g(φ)=cos(φ)，φ(x)=xy
f'(x)=[f'(g)][g'(φ)][φ'(x)]
因此：
d[cos²(xy)]/dx=[2cos(xy)][cos'(xy)][(xy)']
=[2cos(xy)][-sin(xy)](y)
=-2ycos(xy)sin(xy)
=-ysin(2xy)
同样的，有：
d[cos²(xy)]/dy=-xsin(2xy)
解：
cos²(xy)，可以看做是复合函数：
f(g)=g²，g(φ)=cos(φ)，φ(x)=xy
f'(x)=[f'(g)][g'(φ)][φ'(x)]
因此：
d[cos²(xy)]/dx=[2cos(xy)][cos'(xy)][(xy)']
=[2cos(xy)][-sin(xy)](y)
=-2ycos(xy)sin(xy)
=-ysin(2xy)
同样的，有：...
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