当 x > 0 时, 令 u = t√x , 则 t = u/√x, dt = du/√x
f(x,y) = ∫<0, xy> e^(xt^2)dt = ∫<0, yx^(3/2)> e^(u^2)du/√x
= (1/√x)∫<0, yx^(3/2)> e^(u^2)du
∂f/∂x = [(-1/2)/x^(3/2)]∫<0, yx^(3/2)> e^(u^2)du + (3/2)ye^(x^3y^2)
∂^2f/∂x∂y = (-1/2)e^(x^3y^2) + (3/2)e^(x^3y^2) + 3x^3y^2e^(x^3y^2)
x = 1, y = 1 时 , ∂^2f/∂x∂y = 4e
σf/σy=e^(x(xy)²)(xy)'=xe^(x³y²)
σf²/σxσy=e^(x³y²)+3x³y²e^(x³y²)
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