2sin(2+2y-3z)=x+2y-3z,证明δz/δx+δz/δy=1

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2sin(2+2y-3z)=x+2y-3z,证明δz/δx+δz/δy=1
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2sin(2+2y-3z)=x+2y-3z,证明δz/δx+δz/δy=1
2sin(2+2y-3z)=x+2y-3z,证明δz/δx+δz/δy=1

2sin(2+2y-3z)=x+2y-3z,证明δz/δx+δz/δy=1
已知2sin(x+2y-3z)=x+2y-3z,求证∂z/∂x+∂z/∂y=1
【原题有错!2sin(2+2y-3z)应是2sin(x+2y-3z)之误,否则等式不能成立.】
证明：令F(x,y,z)=2sin(x+2y-3z)-x-2y+3z=0
则∂z/∂x=-(∂F/∂x)/(∂F/∂z)=-[2cos(x+2y-3z)-1]/[-6cos(x+2y-3z)+3].(1)
∂z/∂y=-(∂F/∂y)/(∂F/∂z)=-[4cos(x+2y-3z)-2]/[-6cos(x+2y-3z)+3].(2)
(1)+(2)即得∂z/∂x+∂z/∂y=[-6cos(x+2y-3z)+3]/[-6cos(x+2y-3z)+3]=1

令u=x+2y-3z
所以2sin(1+u)=u
所以F(u)=2sin(1+u)-u=0
F'x=(F'u)*(u'x)=2cos(1+u)-1
F'y=(F'u)*(u'y)=[2cos(1+u)-1]*2=4cos(1+u)-2
F'z=(F'u)*(u'z)=[2cos(1+u)-1]*(-3)=-6cos(1+u)+3
z'x=-F'x/F...

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令u=x+2y-3z
所以2sin(1+u)=u
所以F(u)=2sin(1+u)-u=0
F'x=(F'u)*(u'x)=2cos(1+u)-1
F'y=(F'u)*(u'y)=[2cos(1+u)-1]*2=4cos(1+u)-2
F'z=(F'u)*(u'z)=[2cos(1+u)-1]*(-3)=-6cos(1+u)+3
z'x=-F'x/F'z=[2cos(1+u)-1]/[6cos(1+u)-3]
z'y=-F'y/F'z=[4cos(1+u)-2]/[6cos(1+u)-3]
所以
z'x+z'y=[2cos(1+u)-1]/[6cos(1+u)-3] + [4cos(1+u)-2]/[6cos(1+u)-3]=1

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