
设f(u,v)具有二阶连续偏导数,z=f(e^xsiny,x^2+y^2). 计算δ^2z/δx^2 (δ为偏导数符号) 急求解答步
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令e^xsiny=u,x^2+y^2=v
则δz/δx
=δf/δu*δu/δx+δf/δv*δv/δx
=δf/δu*(e^xsiny)+δf/δv*(2x)
δ^2z/δx^2
=δ^2f/δu^2*(e^xsiny)*(e^xsiny)+δ^2f/δuδv*(2x)*(e^xsiny)+δf/δu*(e^xsiny)+δ^2f/δvδu*(e^xsiny)*2x+δ^2f/δv^2*(2x)*(2x)+2δf/δv
=(e^2x*(siny)^2)*δ^2f/δu^2+(e^xsiny)* δf/δu+(4xe^xsiny)*δ^2f/δuδv+4x^2*δ^f/δ^2v+2δf/δv
(f(u,v)具有二阶连续偏导数=>δ^2f/δuδv=δ^2f/δvδu)
则δz/δx
=δf/δu*δu/δx+δf/δv*δv/δx
=δf/δu*(e^xsiny)+δf/δv*(2x)
δ^2z/δx^2
=δ^2f/δu^2*(e^xsiny)*(e^xsiny)+δ^2f/δuδv*(2x)*(e^xsiny)+δf/δu*(e^xsiny)+δ^2f/δvδu*(e^xsiny)*2x+δ^2f/δv^2*(2x)*(2x)+2δf/δv
=(e^2x*(siny)^2)*δ^2f/δu^2+(e^xsiny)* δf/δu+(4xe^xsiny)*δ^2f/δuδv+4x^2*δ^f/δ^2v+2δf/δv
(f(u,v)具有二阶连续偏导数=>δ^2f/δuδv=δ^2f/δvδu)
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