求高手解答z=ln(x^2+y^2)的二阶偏导数?
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z=ln(x²+y²)
∂z/∂x=2x/(x²+y²)
∂²z/∂x²=2(y²-x²)/(x²+y²)²
∂z/∂y=2y/(x²+y²)
∂²z/∂y²=2(x²-y²)/(x²+y²)²
∂²z/∂x∂y=-4xy/(x²+y²)²,9,az/ax=2x/(x^2+y^2) az/ay=2y/(x^2+y^2)
a^2z/ax^2=(2y^2-2x^2)/(x^2+y^2)^4 a^2z/ay^2=(2x^2-2y^2)/(x^2+y^2)^4,0,
∂z/∂x=2x/(x²+y²)
∂²z/∂x²=2(y²-x²)/(x²+y²)²
∂z/∂y=2y/(x²+y²)
∂²z/∂y²=2(x²-y²)/(x²+y²)²
∂²z/∂x∂y=-4xy/(x²+y²)²,9,az/ax=2x/(x^2+y^2) az/ay=2y/(x^2+y^2)
a^2z/ax^2=(2y^2-2x^2)/(x^2+y^2)^4 a^2z/ay^2=(2x^2-2y^2)/(x^2+y^2)^4,0,
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