題目 6. (10 pts) Let g:(0,∞)→Rg:(0,\infty)\to\mathbb{R}g:(0,∞)→R be a twice differentiable function. Assume that g(1)=1, g′(1)=3, g′′(1)=−4.g(1)=1,\ g'(1)=3,\ g''(1)=-4.g(1)=1, g′(1)=3, g′′(1)=−4. Define a real valued function hhh on R3∖{(0,0,0)}\mathbb{R}^3 \setminus \{(0,0,0)\}R3∖{(0,0,0)} by h(x,y,z)=g(x2+y2+z2)h(x,y,z)=g\left(\sqrt{x^2+y^2+z^2}\right)h(x,y,z)=g(x2+y2+z2) Calculate ∂2h∂x2(P)+∂2h∂z2(P)+∂2h∂z2(P)\dfrac{\partial^2 h}{\partial x^2}(P)+\dfrac{\partial^2 h}{\partial z^2}(P)+\dfrac{\partial^2 h}{\partial z^2}(P)∂x2∂2h(P)+∂z2∂2h(P)+∂z2∂2h(P) where P=(23,23,−13)P=\left(\dfrac{2}{3}, \dfrac{2}{3}, -\dfrac{1}{3}\right)P=(32,32,−31). 提醒:官方原題此處有筆誤,實際上應為 ∂2h∂x2(P)+∂2h∂y2(P)+∂2h∂z2(P)\dfrac{\partial^2 h}{\partial x^2}(P)+\dfrac{\partial^2 h}{\partial y^2}(P)+\dfrac{\partial^2 h}{\partial z^2}(P)∂x2∂2h(P)+∂y2∂2h(P)+∂z2∂2h(P)。 解答