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设z=f(exsiny,x2+y2),其中f具有二阶连续偏导数,求
2z
∂x∂y
人气:498 ℃ 时间:2020-06-23 11:50:56
解答

∵z=f(exsiny,x2+y2),
∂z
∂x
=f1•[exsiny]x+f2•[x2+y2]x
=exsinyf′1+2xf′2
进一步得:
2z
∂x∂y
∂y
(
∂z
∂x
)
=[exsinyf'1]y+[2xf′2]y
=ex[cosyf1+siny•
∂f1
∂y
]+2x
∂f2
∂y

=excosyf1+exsiny•[f11excosy+f12•2y]+2x[f21excosy+f22•2y]
=e2xsinycosyf11+2ex(ysiny+xcosy)f12+4xyf22+excosyf1
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