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∂z/∂x=1/√(1-x²y²)*(xy)'=y/√(1-x²y²)=y*(1-x²y²)^(-1/2)
所以∂²z/∂x²=-1/2*y*(1-x²y²)^(-3/2)*(1-x²y²)
=-1/2*y/[(1-x²y²)*√(1-x²y²)](-2xy²)
=xy³/[(1-x²y²)*√(1-x²y²)]
同理
∂²z/∂y²=x³y/[(1-x²y²)*√(1-x²y²)]
所以∂²z/∂x²=-1/2*y*(1-x²y²)^(-3/2)*(1-x²y²)
=-1/2*y/[(1-x²y²)*√(1-x²y²)](-2xy²)
=xy³/[(1-x²y²)*√(1-x²y²)]
同理
∂²z/∂y²=x³y/[(1-x²y²)*√(1-x²y²)]
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