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z=xf(x^2+y^2)的二阶偏导数问题
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z=xf(x^2+y^2),
dz=f(x^2+y^2)dx+xf'(x^2+y^2)(2xdx+2ydy)
dz=[f(x^2+y^2)+2x^2f'(x^2+y^2)]dx+2xyf'(x^2+y^2)dy
dz/dx=f(x^2+y^2)+2x^2f'(x^2+y^2)
d^2z/dxdy=2yf'(x^2+y^2)+2x^2f''(x^2+y^2)2y
=2yf'(x^2+y^2)+4x^2yf''(x^2+y^2).
dz=f(x^2+y^2)dx+xf'(x^2+y^2)(2xdx+2ydy)
dz=[f(x^2+y^2)+2x^2f'(x^2+y^2)]dx+2xyf'(x^2+y^2)dy
dz/dx=f(x^2+y^2)+2x^2f'(x^2+y^2)
d^2z/dxdy=2yf'(x^2+y^2)+2x^2f''(x^2+y^2)2y
=2yf'(x^2+y^2)+4x^2yf''(x^2+y^2).
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