[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)=xyf'(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(...