设函数z=sin(x^2-2y) 求二阶偏导数
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解：
dz/dx=2xcos(x²-2y)
d²z/d²x
=[2xcos(x²-2y)]'
=2{x'cos(x²-2y)+x[cos(x²-2y)]'}
=2[cos(x²-2y)-xsin(x²-2y)2x]
=2[cos(x²-2y)-2x²sin(x²-2y)]
dz/dy=-2cos(x²-2y)
d²z/d²y
=[-2cos(x²-2y)]'
=-2[cos(x²-2y)]'
=-2[-sin(x²-2y)](-2)
=-4sin(x²-2y)
dz/dx=2xcos(x²-2y)
d²z/d²x
=[2xcos(x²-2y)]'
=2{x'cos(x²-2y)+x[cos(x²-2y)]'}
=2[cos(x²-2y)-xsin(x²-2y)2x]
=2[cos(x²-2y)-2x²sin(x²-2y)]
dz/dy=-2cos(x²-2y)
d²z/d²y
=[-2cos(x²-2y)]'
=-2[cos(x²-2y)]'
=-2[-sin(x²-2y)](-2)
=-4sin(x²-2y)
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