培風館『入門微分積分』第4章・問題4.3.3の解答解説です。問題文、解法の要点、途中式を含む解答例を整理しています。
問題4.3.3
（１）$z=\log(x^2+y^2)$
（２）$z=\dfrac{x}{x^2+y^2}$
（３）$z=\tan^{-1} \dfrac{y}{x}$
（４）$z=x^3+xy+y^3$
解法の要点
解答例
$$\begin{align} \varDelta z &= \left(\dfrac{\partial ^2}{\partial x^2}+\dfrac{\partial ^2}{\partial y^2}\right)\log(x^2+y^2) \\ &=\dfrac{\partial ^2}{\partial x^2}\log(x^2+y^2)+\dfrac{\partial ^2}{\partial y^2}\log(x^2+y^2) \\ &=\dfrac{\partial}{\partial x}\left(\dfrac{2x}{x^2+y^2}\right)+\dfrac{\partial}{\partial y}\left(\dfrac{2y}{x^2+y^2}\right) \\ &=\dfrac{2(-x^2+y^2)}{x^2+y^2}+\dfrac{2(x^2-y^2)}{x^2+y^2} \\ &=0 \ \ \cdots \cdots \text{(答)} \end{align}$$
（２）$z=\dfrac{x}{x^2+y^2}$
$$\begin{align} \varDelta z &= \left(\dfrac{\partial ^2}{\partial x^2}+\dfrac{\partial ^2}{\partial y^2}\right)\dfrac{x}{x^2+y^2} \\
&=\dfrac{\partial}{\partial x}\left(\dfrac{-x^2+y^2}{(x^2+y^2)^2}\right)+\dfrac{\partial}{\partial y}\left(\dfrac{-2xy}{(x^2+y^2)^2}\right) \\
&=\dfrac{2 x (x^2 – 3 y^2)}{(x^2 + y^2)^3}-\dfrac{2 x (x^2 – 3 y^2)}{(x^2 + y^2)^3} \\ &=0 \ \ \cdots \cdots \text{(答)} \end{align}$$
（３）$z=\tan^{-1} \dfrac{y}{x}$
$$\begin{align} \varDelta z &= \left(\dfrac{\partial ^2}{\partial x^2}+\dfrac{\partial ^2}{\partial y^2}\right)\tan^{-1} \dfrac{y}{x} \\
&=\dfrac{\partial}{\partial x}\left(\dfrac{-y}{x^2+y^2}\right)+\dfrac{\partial}{\partial y}\left(\dfrac{x}{x^2+y^2}\right) \\
&=\dfrac{2xy}{(x^2 + y^2)^2}-\dfrac{2xy}{(x^2 + y^2)^2} \\ &=0 \ \ \cdots \cdots \text{(答)} \end{align}$$
（４）$z=x^3+xy+y^3$
$$\begin{align} \varDelta z &= \left(\dfrac{\partial ^2}{\partial x^2}+\dfrac{\partial ^2}{\partial y^2}\right)(x^3+xy+y^3) \\
&=\dfrac{\partial}{\partial x}(3x^2+y)+\dfrac{\partial}{\partial y}(x+3y^2) \\
&=6x+6y \\ &=6(x+y) \ \ \cdots \cdots \text{(答)} \end{align}$$