[cos(xy)]的平方,如何分别对x和y求导?

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[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)]的平方,如何分别对x和y求导?
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)

链式法则
[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)]