直接の微分計算など，フツーはやらないことのメモ．
気になったとき（そんなことある？）に確認するためのもの．
直接計算でいかにラクをするかを考えるのはちょっと楽しい．
【関連記事】
1階微分
$r$を$\boldsymbol{x}=(x,y,z)=(x^{1},x^{2},x^{3})$の関数とするとき，$f(r)=e^{ikr}/r$に対して\begin{aligned}
\frac{\partial f}{\partial x^{i}} \bigl(r(\boldsymbol{x})\bigr)
&=\frac{\mathrm{d} f}{\mathrm{d}r} \bigl(r(\boldsymbol{x})\bigr)
\frac{\partial r}{\partial x^{i}} (\boldsymbol{x}) \\
&=\biggl(\frac{ik e^{ikr}}{r}-\frac{e^{ikr}}{r^{2}}\biggr)
\frac{\partial r}{\partial x^{i}} \\
&=\biggl(ik - \frac{1}{r}\biggr) f \frac{\partial r}{\partial x^{i}}
\end{aligned}
となります（途中から変数を略した）．\frac{\partial f}{\partial x^{i}} \bigl(r(\boldsymbol{x})\bigr)
&=\frac{\mathrm{d} f}{\mathrm{d}r} \bigl(r(\boldsymbol{x})\bigr)
\frac{\partial r}{\partial x^{i}} (\boldsymbol{x}) \\
&=\biggl(\frac{ik e^{ikr}}{r}-\frac{e^{ikr}}{r^{2}}\biggr)
\frac{\partial r}{\partial x^{i}} \\
&=\biggl(ik - \frac{1}{r}\biggr) f \frac{\partial r}{\partial x^{i}}
\end{aligned}
特に，$r=\sqrt{x^{2}+y^{2}+z^{2}}$の場合には
\begin{aligned}
\frac{\partial r}{\partial x^{i}}
&=\frac{x^{i}}{r}
\end{aligned}
です．\frac{\partial r}{\partial x^{i}}
&=\frac{x^{i}}{r}
\end{aligned}
2階微分
同一変数の場合
後で見るように，簡単に一般化できるのですが，まずは同一変数による2階微分を計算します．$x^{i}$の添字$i$も略して同じ変数による2階の偏微分を計算すると，積の微分法から
\begin{aligned}
\frac{\partial^{2} f}{\partial x^{2}}
&=\frac{1}{r^{2}} f \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
+\biggl(ik - \frac{1}{r}\biggr)^{2} f \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
+ \biggl(ik - \frac{1}{r}\biggr) f \frac{\partial^{2} r}{\partial x^{2}} \\
&= f\biggl\{ -\biggl(\frac{\partial r}{\partial x}\biggr)^{2}
\biggl[k^{2} + \frac{2}{r} \biggl(ik - \frac{1}{r}\biggr) \biggr]
+ \biggl(ik - \frac{1}{r}\biggr) \frac{\partial^{2} r}{\partial x^{2}}
\biggr\} \\
&= f\biggl\{\biggl(ik - \frac{1}{r}\biggr)
\biggl[-\frac{2}{r} \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
+\frac{\partial^{2} r}{\partial x^{2}} \biggr]
-k^{2} \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
\biggr\} \\
&= f\biggl\{\frac{1}{r} \biggl(ik - \frac{1}{r}\biggr)
\biggl[-3 \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
+ \frac{\partial}{\partial x}\biggl(r\frac{\partial r}{\partial x}\biggr) \biggr]
-k^{2} \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
\biggr\}
\end{aligned}
となります．\frac{\partial^{2} f}{\partial x^{2}}
&=\frac{1}{r^{2}} f \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
+\biggl(ik - \frac{1}{r}\biggr)^{2} f \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
+ \biggl(ik - \frac{1}{r}\biggr) f \frac{\partial^{2} r}{\partial x^{2}} \\
&= f\biggl\{ -\biggl(\frac{\partial r}{\partial x}\biggr)^{2}
\biggl[k^{2} + \frac{2}{r} \biggl(ik - \frac{1}{r}\biggr) \biggr]
+ \biggl(ik - \frac{1}{r}\biggr) \frac{\partial^{2} r}{\partial x^{2}}
\biggr\} \\
&= f\biggl\{\biggl(ik - \frac{1}{r}\biggr)
\biggl[-\frac{2}{r} \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
+\frac{\partial^{2} r}{\partial x^{2}} \biggr]
-k^{2} \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
\biggr\} \\
&= f\biggl\{\frac{1}{r} \biggl(ik - \frac{1}{r}\biggr)
\biggl[-3 \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
+ \frac{\partial}{\partial x}\biggl(r\frac{\partial r}{\partial x}\biggr) \biggr]
-k^{2} \biggl(\frac{\partial r}{\partial x}\biggr)^{2}
\biggr\}
\end{aligned}
特に，$r=\sqrt{x^{2}+y^{2}+z^{2}}$の場合には
\begin{aligned}
\frac{\partial r}{\partial x}
&=\frac{x}{r}
\end{aligned}
なので，\frac{\partial r}{\partial x}
&=\frac{x}{r}
\end{aligned}
\begin{aligned}
\frac{\partial^{2} f}{\partial x^{2}}
&= f\biggl[\frac{1}{r} \biggl(ik - \frac{1}{r}\biggr) \biggl(1-3\frac{x^{2}}{r^{2}}\biggr)
-k^{2} \frac{x^{2}}{r^{2}}
\biggr]
\end{aligned}
となります．\frac{\partial^{2} f}{\partial x^{2}}
&= f\biggl[\frac{1}{r} \biggl(ik - \frac{1}{r}\biggr) \biggl(1-3\frac{x^{2}}{r^{2}}\biggr)
-k^{2} \frac{x^{2}}{r^{2}}
\biggr]
\end{aligned}
ヘルムホルツ方程式
$x^{2}+y^{2}+z^{2} = r^{2}$に気をつければ\begin{aligned}
\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} +\frac{\partial^{2} f}{\partial z^{2}}
=-k^{2}f
\end{aligned}
が導かれます．\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} +\frac{\partial^{2} f}{\partial z^{2}}
=-k^{2}f
\end{aligned}
※ラプラシアンの極座標表示から示すほうが遥かにラクです．
ラプラシアン（極座標・円筒座標）の計算はヤコビアンを使うと簡単 - Notes_JP
一般化
同一変数の制限を外します．ほとんど同じように計算できます．\begin{aligned}
\frac{\partial^{2} f}{\partial x \partial y}
&=\frac{1}{r^{2}} f \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
+\biggl(ik - \frac{1}{r}\biggr)^{2} f \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
+ \biggl(ik - \frac{1}{r}\biggr) f \frac{\partial^{2} r}{\partial x \partial y} \\
&= f\biggl\{ -\frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
\biggl[k^{2} + \frac{2}{r} \biggl(ik - \frac{1}{r}\biggr) \biggr]
+ \biggl(ik - \frac{1}{r}\biggr) \frac{\partial^{2} r}{\partial x \partial y}
\biggr\} \\
&= f\biggl\{\biggl(ik - \frac{1}{r}\biggr)
\biggl[-\frac{2}{r} \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
+\frac{\partial^{2} r}{\partial x \partial y} \biggr]
-k^{2} \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
\biggr\} \\
&= f\biggl\{\frac{1}{r} \biggl(ik - \frac{1}{r}\biggr)
\biggl[-3 \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
+ \frac{\partial}{\partial y}\biggl(r\frac{\partial r}{\partial x}\biggr) \biggr]
-k^{2} \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
\biggr\}
\end{aligned}
\frac{\partial^{2} f}{\partial x \partial y}
&=\frac{1}{r^{2}} f \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
+\biggl(ik - \frac{1}{r}\biggr)^{2} f \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
+ \biggl(ik - \frac{1}{r}\biggr) f \frac{\partial^{2} r}{\partial x \partial y} \\
&= f\biggl\{ -\frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
\biggl[k^{2} + \frac{2}{r} \biggl(ik - \frac{1}{r}\biggr) \biggr]
+ \biggl(ik - \frac{1}{r}\biggr) \frac{\partial^{2} r}{\partial x \partial y}
\biggr\} \\
&= f\biggl\{\biggl(ik - \frac{1}{r}\biggr)
\biggl[-\frac{2}{r} \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
+\frac{\partial^{2} r}{\partial x \partial y} \biggr]
-k^{2} \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
\biggr\} \\
&= f\biggl\{\frac{1}{r} \biggl(ik - \frac{1}{r}\biggr)
\biggl[-3 \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
+ \frac{\partial}{\partial y}\biggl(r\frac{\partial r}{\partial x}\biggr) \biggr]
-k^{2} \frac{\partial r}{\partial x} \frac{\partial r}{\partial y}
\biggr\}
\end{aligned}
よって，$r=\sqrt{x^{2}+y^{2}+z^{2}}$の場合に$x=x^{i}, y=x^{j}$として計算すると，
\begin{aligned}
\frac{\partial^{2} f}{\partial x^{i} \partial x^{j}}
&= f\biggl[\frac{1}{r} \biggl(ik - \frac{1}{r}\biggr)
\biggl(-3 \frac{x^{i}}{r} \frac{x^{j}}{r}
+ \frac{\partial x^{i}}{\partial x^{j}} \biggr)
-k^{2} \frac{x^{i}}{r} \frac{x^{j}}{r}
\biggr] \\
&= f\biggl[\frac{1}{r} \biggl(ik - \frac{1}{r}\biggr)
\biggl(\delta^{i}_{~j}
-3 \frac{x^{i}x^{j}}{r^{2}} \biggr)
-k^{2} \frac{x^{i}x^{j}}{r^{2}}
\biggr]
\end{aligned}
となります．
\frac{\partial^{2} f}{\partial x^{i} \partial x^{j}}
&= f\biggl[\frac{1}{r} \biggl(ik - \frac{1}{r}\biggr)
\biggl(-3 \frac{x^{i}}{r} \frac{x^{j}}{r}
+ \frac{\partial x^{i}}{\partial x^{j}} \biggr)
-k^{2} \frac{x^{i}}{r} \frac{x^{j}}{r}
\biggr] \\
&= f\biggl[\frac{1}{r} \biggl(ik - \frac{1}{r}\biggr)
\biggl(\delta^{i}_{~j}
-3 \frac{x^{i}x^{j}}{r^{2}} \biggr)
-k^{2} \frac{x^{i}x^{j}}{r^{2}}
\biggr]
\end{aligned}
