練習問題メモ 03#
和達三樹. (2019). 微分積分. 第5章 練習問題
[1]#
[1] 次の関数は原点 \((0,0)\) を除いて連続である. 原点でも連続にできるか? 直線 \(y\) \(=m x\) に沿って原点に近づくとして調べよ.
\[
\text { (1) } \quad
z=\frac{e^{x^2+y^2}-1}{x^2+y^2}
\]
\[
z=\frac{e^{x^2+m^2x^2}-1}{x^2+m^2x^2}
=\frac{e^{(1+m^2)x^2}-1}{(1+m^2)x^2}
\]
指数関数のテイラー展開
\[
exp(ax) = 1 + ax + \frac{(ax)^2}{2!} + \frac{(ax)^3}{3!} + \cdots
\]
により、\(x\)が小さい時
\[
e^{a x^2} = 1 + a x^2
\]
となるため
\[
z = \frac{e^{(1+m^2)x^2}-1}{(1+m^2)x^2}
= \frac{1 + (1+m^2) x^2 - 1}{(1+m^2)x^2}
= \frac{(1+m^2) x^2}{(1+m^2)x^2} = 1
\]
となり、\(m\)の値の取り方（0への近づき方）によらず1に収束する
よって連続にできる
\[\begin{split}
z = \begin{cases}
\frac{e^{x^2+y^2}-1}{x^2+y^2} & ((x,y) \neq (0,0)のとき)\\
1 & ((x,y) = (0,0)のとき)\\
\end{cases}
\end{split}\]
Show code cell source
import numpy as np
import matplotlib.pyplot as plt
a = 1
x = np.linspace(-1, 1, 100)
plt.plot(x, np.exp(a * x**2))
plt.xlabel("x")
plt.ylabel(f"$\exp(a x^2), a=1$")
plt.title(f"$\exp(a x^2), a=1$")
plt.show()
\[
\text { (2) } \quad
z=\frac{x^2 y^2}{x^4+y^4}
\]
\[
\lim_{x\to 0} z
= \lim_{x\to 0} \frac{x^4 m^2}{x^4 + m^4 x^4}
= \lim_{x\to 0} \frac{m^2}{1 + m^4}
= \frac{m^2}{1 + m^4}
\]
\(m\)の値によって異なる値をとるため、原点への近づけ方によって異なる値になってしまう。よって極限は存在せず、連続にできない。
[2]#
[2] 次の関数 \(f(x, y)\) の \(f_x, f_y, f_{x x}, f_{y y}, f_{x y}, f_{y x}\) を計算せよ.
(1) \(f(x, y)=x^3-x^2 y+x y^2-y^3+1\)
(2) \(f(x, y)=x^2 \cos y-y^2 \cos x\)
(3) \(f(x, y)=\arctan x^2 y\)
(1) \(f(x, y)=x^3-x^2 y+x y^2-y^3+1\)
\[\begin{split}
f_x = 3x^2 - 2xy + y^2\\
f_y = 3y^2 + 2xy - x^2\\
f_{xx} = 6x - 2y\\
f_{yy} = 6y + 2x\\
f_{xy} = - 2x + 2y\\
f_{yx} = - 2x + 2y\\
\end{split}\]
(2) \(f(x, y)=x^2 \cos y-y^2 \cos x\)
\[\begin{split}
f_x = 2x \cos y + y^2 \sin x\\
f_y = - x^2 \sin y - 2y \cos x\\
f_{xx} = 2 \cos y + y^2 \cos x\\
f_{yy} = - x^2 \cos y - 2 \cos x\\
f_{xy} = - 2x \sin y + 2y \sin x\\
f_{yx} = - 2x \sin y + 2y \sin x\\
\end{split}\]
(3) \(f(x, y)=\arctan x^2 y\)
\((\arctan x)' = 1/(1+x^2)\)なので
\[\begin{split}
\begin{align}
f_x &= \frac{1}{1+x^4 y^2} 2xy\\
f_y &= \frac{1}{1+x^4 y^2} x^2\\
f_{xx} &= -\frac{(1+x^4 y^2)'}{(1+x^4 y^2)^2} 2xy + \frac{1}{1+x^4 y^2} (2xy)' \quad (積の微分公式と商の微分公式）\\
&= -\frac{4x^3 y^2}{(1+x^4 y^2)^2} 2xy + \frac{1}{1+x^4 y^2} 2y \\
&= -\frac{8x^4 y^3}{(1+x^4 y^2)^2} + \frac{2y}{1+x^4 y^2} \\
&= -\frac{8x^4 y^3}{(1+x^4 y^2)^2} + \frac{2y(1+x^4 y^2)}{(1+x^4 y^2)^2} \\
&= \frac{2y+ 2x^4 y^2 - 8x^4 y^3}{(1+x^4 y^2)^2} \\
&= \frac{2y - 6x^4 y^3}{(1+x^4 y^2)^2} \\
&= \frac{2y(1 - 3x^4 y^2)}{(1+x^4 y^2)^2} \\
f_{yy} &=-\frac{(1+x^4 y^2)'}{(1+x^4 y^2)^2} x^2 + \frac{1}{1+x^4 y^2} (x^2)' \quad (積の微分公式と商の微分公式）\\
&=-\frac{2 x^6 y}{(1+x^4 y^2)^2}\\
f_{xy} &= \frac{(2xy)' (1 + x^4 y^2) - 2xy (1 + x^4 y^2)'}{(1 + x^4 y^2)^2}\\
&= \frac{2x (1 + x^4 y^2) - 4x^5 y^2}{(1 + x^4 y^2)^2}\\
&= \frac{2x + 2x^5 y^2 - 4x^5 y^2}{(1 + x^4 y^2)^2}\\
&= \frac{2x - 2x^5 y^2}{(1 + x^4 y^2)^2}\\
&= \frac{2x(1 - x^4 y^2)}{(1 + x^4 y^2)^2}\\
\\
f_{yx} &= \frac{(x^2)' (1 + x^4 y^2) - x^2 (1 + x^4 y^2)'}{(1 + x^4 y^2)^2}\\
&= \frac{2x (1 + x^4 y^2) - x^2 \times 4x^3 y^2}{(1 + x^4 y^2)^2}\\
&= \frac{2x + 2x^5 y^2 - 4x^5 y^2}{(1 + x^4 y^2)^2}\\
&= \frac{2x - 2x^5 y^2}{(1 + x^4 y^2)^2}\\
&= \frac{2x(1 - x^4 y^2)}{(1 + x^4 y^2)^2}\\
\end{align}
\end{split}\]
import sympy as sp
x, y = sp.symbols('x y')
f = sp.atan(x**2 * y)
df = sp.diff(f, x)
sp.diff(df, x)
\[\displaystyle - \frac{8 x^{4} y^{3}}{\left(x^{4} y^{2} + 1\right)^{2}} + \frac{2 y}{x^{4} y^{2} + 1}\]
import sympy as sp
x, y = sp.symbols('x y')
f = sp.atan(x**2 * y)
df = sp.diff(f, x)
sp.diff(df, x)
\[\displaystyle - \frac{8 x^{4} y^{3}}{\left(x^{4} y^{2} + 1\right)^{2}} + \frac{2 y}{x^{4} y^{2} + 1}\]
[3]#
[3] 次の関数の偏導関数 \(\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}\) を求めよ.
(1) \(z=x^2+4 x y+y^3\)
(2) \(z=\frac{x}{y^2}-\frac{y}{x^2}\)
(3) \(z=\sin (2 x+1) \cos \left(y^2+4\right)\)
(4) \(3 x^2+4 y^2-5 z^2=20\)
(5) \(\sin x y+\sin y z+\sin x z=1\)
(1) \(z=x^2+4 x y+y^3\)
\[\begin{split}
\frac{\partial z}{\partial x} = 2x + 4y
\\
\frac{\partial z}{\partial y} = 3y^2 + 4x
\end{split}\]
(2) \(z=\frac{x}{y^2}-\frac{y}{x^2}\)
\[\begin{split}
\begin{align}
\frac{\partial z}{\partial x}
&= \frac{1}{y^2} (x') - y \left(\frac{1}{x^2} \right)'\\
&= \frac{1}{y^2} (x') + y \frac{(x^2)'}{(x^2)^2} \quad (商の微分公式)\\
&= \frac{1}{y^2} + \frac{2y}{x^3}\\
\frac{\partial z}{\partial y}
&= \left(\frac{1}{y^2}\right)' x - \frac{1}{x^2} (y)'\\
&= -\frac{2y}{y^4} x - \frac{1}{x^2}\\
&= -\frac{2x}{y^3} - \frac{1}{x^2}\\
\end{align}
\end{split}\]
(3) \(z=\sin (2 x+1) \cos \left(y^2+4\right)\)
\[\begin{split}
\begin{align}
\frac{\partial z}{\partial x}
&= (\sin(2 x+1))' \cos(y^2+4)\\
&= 2 \cos(2x+1) \cos(y^2 + 4)\\
\\
\frac{\partial z}{\partial y}
&= \sin(2 x+1) (\cos(y^2+4))'\\
&= -2y \sin(2x+1) \sin(y^2 + 4)\\
\end{align}
\end{split}\]
import sympy as sp
x, y = sp.symbols('x y')
f = sp.sin(2*x + 1) * sp.cos(y**2 + 4)
sp.diff(f, y)
\[\displaystyle - 2 y \sin{\left(2 x + 1 \right)} \sin{\left(y^{2} + 4 \right)}\]
(4) \(3 x^2+4 y^2-5 z^2=20\)
\[
F(x,y,z) := 3x^2 + 4y^2 - 5z^2 - 20 = 0
\]
とおく。
\[
\frac{\partial F}{\partial x} = 6x, \quad
\frac{\partial F}{\partial y} = 8y, \quad
\frac{\partial F}{\partial z} = -10z
\]
であり、\(\frac{\partial F}{\partial z} \neq 0\)のため、定理
\[
\frac{\partial z}{\partial x} = -\frac{\partial F/\partial x}{\partial F / \partial z} , \quad
\frac{\partial z}{\partial y} = -\frac{\partial F/\partial y}{\partial F / \partial z}
\]
より
\[\begin{split}
\begin{align}
\frac{\partial z}{\partial x}
&= \frac{\partial F/\partial x}{\partial F / \partial z}
= \frac{6x}{10z}
= \frac{3x}{5z}
\\
\frac{\partial z}{\partial y}
&= \frac{\partial F/\partial y}{\partial F / \partial z}
= \frac{8y}{10z}
= \frac{4y}{5z}
\end{align}
\end{split}\]
(5) \(\sin x y+\sin y z+\sin x z=1\)
\[
F(x,y,z) := \sin x y+\sin y z+\sin x z - 1 = 0
\]
とおく。
\[\begin{split}
\begin{align}
\frac{\partial F}{\partial x} &= y \cos xy + z \cos xz\\
\frac{\partial F}{\partial y} &= x \cos xy + z \cos yz\\
\frac{\partial F}{\partial z} &= y \cos yz + x \cos xz\\
\end{align}
\end{split}\]
であり、\(\frac{\partial F}{\partial z} \neq 0\)のため、定理より
\[\begin{split}
\begin{align}
\frac{\partial z}{\partial x}
&= -\frac{\partial F/\partial x}{\partial F / \partial z}
= -\frac{y \cos xy + z \cos xz}{y \cos yz + x \cos xz}
\\
\frac{\partial z}{\partial y}
&= -\frac{\partial F/\partial y}{\partial F / \partial z}
= -\frac{x \cos xy + z \cos yz}{y \cos yz + x \cos xz}
\end{align}
\end{split}\]
import sympy as sp
x, y, z = sp.symbols('x y z')
f = sp.sin(x*y) + sp.sin(y*z) + sp.sin(x*z) - 1
sp.diff(f, y)
\[\displaystyle x \cos{\left(x y \right)} + z \cos{\left(y z \right)}\]
[4]#
[4] \(x=\rho \cos \phi, y=\rho \sin \phi\) のとき, 次のことを証明せよ.
(1) \(x \frac{\partial f}{\partial y}-y \frac{\partial f}{\partial x}=0\) ならば, \(f(x, y)\) は \(\rho\) だけの関数である.
(2) \(x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=0\) ならば, \(f(x, y)\) は \(\phi\) だけの関数である.
(3) \(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial f}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}\)
問題略解
\[\begin{split}
\begin{aligned}
& \rho=\sqrt{x^2+y^2}, \quad \phi = \arctan (y / x)\\
& \frac{\partial \rho}{\partial x} = \frac{x}{\rho}=\cos \phi, \quad \frac{\partial \rho}{\partial y} = \frac{y}{\rho}=\sin \phi \\
& \frac{\partial \phi}{\partial x} = -\frac{y}{\rho^2}= -\frac{\sin \phi}{\rho}, \quad \frac{\partial \phi}{\partial y}=\frac{x}{\rho^2}=\frac{\cos \phi}{\rho} \\
& \frac{\partial f}{\partial x}=\frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x}=\cos \phi \frac{\partial f}{\partial \rho}-\frac{\sin \phi}{\rho} \frac{\partial f}{\partial \phi} \\
& \frac{\partial f}{\partial y}=\frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial y}=\sin \phi \frac{\partial f}{\partial \rho}+\frac{\cos \phi}{\rho} \frac{\partial f}{\partial \phi}
\end{aligned}
\end{split}\]
(1) \(x(\partial f / \partial y)-y(\partial f / \partial x)=\partial f / \partial \phi=0\). よって, \(f\) は \(\phi\) によらず, \(\rho\) だけの関数.
(2) \(x(\partial f / \partial x)+y(\partial f / \partial y)=\rho(\partial f / \partial \rho)=0\). よって, \(f\) は \(\rho\) によらず, \(\phi\) だけの関数.
(3)
\[\begin{split}
\begin{aligned}
\frac{\partial^2 f}{\partial x^2}= & \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \rho}{\partial x}+\frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \phi}{\partial x} \\
= & \frac{\partial}{\partial \rho}\left(\cos \phi \frac{\partial f}{\partial \rho}-\frac{\sin \phi}{\rho} \frac{\partial f}{\partial \phi}\right) \cdot \cos \phi \\
& +\frac{\partial}{\partial \phi}\left(\cos \phi \frac{\partial f}{\partial \rho}-\frac{\sin \phi}{\rho} \frac{\partial f}{\partial \phi}\right)\left(-\frac{\sin \phi}{\rho}\right) \\
= & \cos ^2 \phi \frac{\partial^2 f}{\partial \rho^2}+\frac{2 \sin \phi \cos \phi}{\rho^2} \frac{\partial f}{\partial \phi}-\frac{2 \sin \phi \cos \phi}{\rho} \frac{\partial^2 f}{\partial \rho \partial \phi} \\
& +\frac{\sin ^2 \phi}{\rho} \frac{\partial f}{\partial \rho}+\frac{\sin ^2 \phi}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}
\end{aligned}
\end{split}\]
同様にして,
\[\begin{split}
\begin{aligned}
\frac{\partial^2 f}{\partial y^2}= & \sin ^2 \phi \frac{\partial^2 f}{\partial \rho^2}-\frac{2 \sin \phi \cos \phi}{\rho^2} \frac{\partial f}{\partial \phi}+\frac{2 \sin \phi \cos \phi}{\rho} \frac{\partial^2 f}{\partial \rho \partial \phi} \\
& +\frac{\cos ^2 \phi}{\rho} \frac{\partial f}{\partial \rho}+\frac{\cos ^2 \phi}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}
\end{aligned}
\end{split}\]
よって,
\[
\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial f}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}
\]
答例
(2) \(x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=0\) ならば, \(f(x, y)\) は \(\phi\) だけの関数である.
\(x=\rho \cos \phi, y=\rho \sin \phi\) 。
合成関数の微分（p.121）より
\[
\frac{\partial f}{\partial \rho}
=
\frac{\partial f}{\partial x}
\frac{\partial x}{\partial \rho}
+
\frac{\partial f}{\partial y}
\frac{\partial y}{\partial \rho}
\]
ここで
\[
\frac{\partial x}{\partial \rho} = \cos \phi = \frac{x}{\rho}
,\quad
\frac{\partial y}{\partial \rho} = \sin \phi = \frac{y}{\rho}
\]
なので、
\[
\frac{\partial f}{\partial \rho} =
\frac{x}{\rho}
\frac{\partial f}{\partial x}
+
\frac{y}{\rho}
\frac{\partial f}{\partial y}
=
\frac{1}{\rho}
\left(
x
\frac{\partial f}{\partial x}
+
y
\frac{\partial f}{\partial y}
\right)
\]
もし
\[
\left(
x
\frac{\partial f}{\partial x}
+
y
\frac{\partial f}{\partial y}
\right)
\]
がゼロならば\(\frac{\partial f}{\partial \rho}=0\)であり、導関数がゼロということはすなわち\(f\)は\(\rho\)に依存せず\(\phi\)のみの関数
合成関数の微分の定理（p.122）
\(z=f(x, y)\) において \(x\)と\(y\)は変数\(u\)と\(v\)に依存して、
\[
x=g(u, v), \quad y=h(u, v)
\]
とすると
\[
d z=\frac{\partial z}{\partial x} d x+\frac{\partial z}{\partial y} d y, \quad d x=\frac{\partial x}{\partial u} d u+\frac{\partial x}{\partial v} d v, \quad d y=\frac{\partial y}{\partial u} d u+\frac{\partial y}{\partial v} d v
\]
あとの2つの式を最初の式に代入して
\[\begin{split}
\begin{aligned}
d z & =\frac{\partial z}{\partial x}\left(\frac{\partial x}{\partial u} d u+\frac{\partial x}{\partial v} d v\right)+\frac{\partial z}{\partial y}\left(\frac{\partial y}{\partial u} d u+\frac{\partial y}{\partial v} d v\right) \\
& =\left(\frac{\partial z}{\partial x} \frac{\partial x}{\partial u}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial u}\right) d u+\left(\frac{\partial z}{\partial x} \frac{\partial x}{\partial v}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial v}\right) d v
\end{aligned}
\end{split}\]
\(z\)は\(u\)と\(v\)の関数とみなせるから
\[
d z=\frac{\partial z}{\partial u} d u+\frac{\partial z}{\partial v} d v
\]
\(du\)と\(dv\)の係数（偏導関数の部分）をそれぞれ等しいとおくと
\[\begin{split}
\begin{aligned}
& \frac{\partial z}{\partial u}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial u}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial u} \\
& \frac{\partial z}{\partial v}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial v}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial v}
\end{aligned}
\end{split}\]
\(x=\rho \cos \phi, y=\rho \sin \phi\)
(1) \(x \frac{\partial f}{\partial y}-y \frac{\partial f}{\partial x}=0\) ならば, \(f(x, y)\) は \(\rho\) だけの関数である.
定理より
\[
\frac{\partial f}{\partial \phi}
=
\frac{\partial f}{\partial x}
\frac{\partial x}{\partial \phi}
+
\frac{\partial f}{\partial y}
\frac{\partial y}{\partial \phi}
\]
ここで
\[
\frac{\partial x}{\partial \phi}
= - \rho \sin \phi = -y
,\quad
\frac{\partial y}{\partial \phi}
= \rho \cos \phi = x
\]
であるから
\[
\frac{\partial f}{\partial \phi}
=
x
\frac{\partial f}{\partial y}
-
y
\frac{\partial f}{\partial x}
\]
となる
よって\(x \frac{\partial f}{\partial y} -y \frac{\partial f}{\partial x} = 0\)なら、\(f\)は\(\phi\)で微分したとき変化がない定数ということなので、\(f\)は\(\rho\)だけの関数
(3) \(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial f}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}\)
考え方：2次の合成関数の偏導関数を求めていく
\[\begin{split}
\begin{aligned}
\frac{\partial^2 f}{\partial x^2}
&= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)\\
&= \frac{\partial}{\partial x}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x} \right)\\
&= \frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \rho}{\partial x}+\frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \phi}{\partial x} \\
\end{aligned}
\end{split}\]
ここで\(\frac{\partial \rho}{\partial x}\)や\(\frac{\partial \phi}{\partial x}\)を求めるために、\(\rho, \phi\)の関数型とその導関数を求める必要がある
Step 1. \(\rho, \phi\)の関数型の同定：
\[\begin{split}
\begin{aligned}
x^2 + y^2
&= \rho^2 \cos^2 \phi + \rho^2 \sin^2 \phi\\
&= \rho^2 \underbrace{ (\cos^2 \phi + \sin^2 \phi) }_{=1}
= \rho^2
\quad
&\to \rho = \sqrt{x^2 + y^2}
\\
\frac{y}{x} &= \frac{\rho \sin \phi}{\rho \cos \phi} = \tan \phi
\quad
&\to \phi = \arctan(\tan \phi) = \arctan \left(\frac{y}{x} \right)
\end{aligned}
\end{split}\]
よって
\[
\rho=\sqrt{x^2+y^2}, \quad \phi=\tan ^{-1}\left(\frac{y}{x}\right)
\]
Step 2. \(\rho, \phi\)の偏導関数
\[
\frac{\partial \rho}{\partial x}=\frac{x}{\sqrt{x^2+y^2}} = \frac{x}{\rho} =\cos \phi
, \quad
\frac{\partial \rho}{\partial y}=\frac{y}{\sqrt{x^2+y^2}} = \frac{y}{\rho} =\sin \phi
\]
\[
\frac{\partial \phi}{\partial x}=-\frac{y}{x^2+y^2}=-\frac{\sin \phi}{\rho}
, \quad
\frac{\partial \phi}{\partial y}=\frac{x}{x^2+y^2}=\frac{\cos \phi}{\rho}
\]
導出
\[\begin{split}
\begin{aligned}
\frac{\partial \phi}{\partial x}
&= \frac{1}{1 + (y/x)^2}\cdot (-\frac{x'}{x^2} y)\\
&= -\frac{1}{1 + \frac{y^2}{x^2}} \frac{y}{x^2}\\
&= -\frac{1}{\frac{x^2 + y^2}{x^2}} \frac{y}{x^2}\\
&= -\frac{x^2}{x^2 + y^2} \frac{y}{x^2}\\
&= -\frac{y}{x^2+y^2}\\
&= -\frac{\sin \phi}{\rho}\\
\\
\frac{\partial \phi}{\partial y}
&= \frac{1}{1 + (y/x)^2}\cdot \frac{1}{x}\\
&= \frac{x}{x^2 + y^2}\\
&= \frac{\cos \phi}{\rho}\\
\end{aligned}
\end{split}\]
Step 3. \(f\)の偏導関数
合成関数の偏微分
\[\begin{split}
\begin{gathered}
\frac{\partial f}{\partial x}=\frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x} \\
\frac{\partial f}{\partial y}=\frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial y}
\end{gathered}
\end{split}\]
に代入すると
\[\begin{split}
\begin{aligned}
& \frac{\partial f}{\partial x}=\frac{\partial f}{\partial \rho} \cos \phi-\frac{\partial f}{\partial \phi} \frac{\sin \phi}{\rho} \\
& \frac{\partial f}{\partial y}=\frac{\partial f}{\partial \rho} \sin \phi+\frac{\partial f}{\partial \phi} \frac{\cos \phi}{\rho}
\end{aligned}
\end{split}\]
Step 4. \(f\)の2次の偏導関数
\[\begin{split}
\begin{aligned}
\frac{\partial^2 f}{\partial x^2}
&= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)\\
&= \frac{\partial}{\partial x}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x} \right)\\
&= \frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \rho}{\partial x}+\frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \phi}{\partial x} \\
&= \frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial x}\right) \cdot \cos \phi + \frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial x}\right) \cdot \left( -\frac{\sin \phi}{\rho} \right) \\
&= \frac{\partial}{\partial \rho}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x} \right) \cdot \cos \phi + \frac{\partial}{\partial \phi}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x} \right) \cdot \left( -\frac{\sin \phi}{\rho} \right) \\
&= \frac{\partial}{\partial \rho}\left(\cos \phi \frac{\partial f}{\partial \rho}-\frac{\sin \phi}{\rho} \frac{\partial f}{\partial \phi}\right) \cdot \cos \phi
+ \frac{\partial}{\partial \phi}\left(\cos \phi \frac{\partial f}{\partial \rho}-\frac{\sin \phi}{\rho} \frac{\partial f}{\partial \phi}\right)\left(-\frac{\sin \phi}{\rho}\right) \\
&= \frac{\partial}{\partial \rho}\left(\cos^2 \phi \frac{\partial f}{\partial \rho}
\right)
- \frac{\partial}{\partial \rho} \left( \frac{\sin \phi \cos \phi}{\rho} \frac{\partial f}{\partial \phi}\right)
- \frac{\partial}{\partial \phi} \left( \frac{\sin \phi \cos \phi}{\rho} \frac{\partial f}{\partial \rho} \right)
+ \frac{\partial}{\partial \phi} \left( \frac{\sin^2 \phi}{\rho^2} \frac{\partial f}{\partial \phi}\right) \\
&= \cos^2 \phi \frac{\partial^2 f}{\partial \rho^2}
- \frac{\partial}{\partial \rho} \left( \frac{\sin \phi \cos\phi}{\rho} \frac{\partial f}{\partial \phi}\right)
- \frac{\partial}{\partial \phi} \left( \frac{\sin \phi \cos \phi}{\rho} \frac{\partial f}{\partial \rho} \right)
+ \frac{\sin^2 \phi}{\rho^2} \frac{\partial^2 f}{\partial \phi^2} \\
&= \cos ^2 \phi \frac{\partial^2 f}{\partial \rho^2}+\frac{2 \sin \phi \cos \phi}{\rho^2} \frac{\partial f}{\partial \phi}-\frac{2 \sin \phi \cos \phi}{\rho} \frac{\partial^2 f}{\partial \rho \partial \phi}
+\frac{\sin ^2 \phi}{\rho} \frac{\partial f}{\partial \rho}+\frac{\sin ^2 \phi}{\rho^2} \frac{\partial^2 f}{\partial \phi^2} \\
\\
\end{aligned}
\end{split}\]
最後の2行がつながらないが、うまくやればいけそう？
\[\begin{split}
\begin{aligned}
\frac{\partial^2 f}{\partial y^2}
&= \frac{\partial }{\partial y} \left( \frac{\partial f}{\partial y} \right)\\
&= \frac{\partial }{\partial y} \left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial y} \right)\\
&= \frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial y}\right) \cdot \frac{\partial \rho}{\partial y}+\frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial y}\right) \cdot \frac{\partial \phi}{\partial y} \\
&= \frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial y}\right) \cdot \sin \phi + \frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial y}\right) \cdot \frac{\cos \phi}{\rho} \\
&= \frac{\partial}{\partial \rho}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial y} \right) \cdot \sin \phi + \frac{\partial}{\partial \phi}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial y} \right) \cdot \frac{\cos \phi}{\rho} \\
&= \frac{\partial}{\partial \rho} \left( \frac{\partial f}{\partial \rho} \sin \phi +\frac{\partial f}{\partial \phi} \frac{\cos \phi}{\rho} \right) \cdot \sin \phi
+ \frac{\partial}{\partial \phi} \left( \frac{\partial f}{\partial \rho} \sin \phi +\frac{\partial f}{\partial \phi} \frac{\cos \phi}{\rho} \right) \cdot \frac{\cos \phi}{\rho} \\
&= \sin ^2 \phi \frac{\partial^2 f}{\partial \rho^2}-\frac{2 \sin \phi \cos \phi}{\rho^2} \frac{\partial f}{\partial \phi}+\frac{2 \sin \phi \cos \phi}{\rho} \frac{\partial^2 f}{\partial \rho \partial \phi}
+ \frac{\cos ^2 \phi}{\rho} \frac{\partial f}{\partial \rho}+\frac{\cos ^2 \phi}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}
\end{aligned}
\end{split}\]
Step 5. \(f\)の2次の偏導関数の和を求める
\[
\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial \rho^2}\left(\cos ^2 \phi+\sin ^2 \phi\right)+\frac{\partial^2 f}{\partial \phi^2}\left(\frac{\sin ^2 \phi+\cos ^2 \phi}{\rho^2}\right)+\frac{1}{\rho} \frac{\partial f}{\partial \rho}
\]
\(\cos ^2 \phi+\sin ^2 \phi=1\)のため
\[
\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial f}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}
\]
ということらしい
[5]#
[5] 3 つの変数 \(x, y, z\) の間に \(F(x, y, z)=0\) の関係がある. 次のことを示せ.
(1) \(\left(\frac{\partial y}{\partial x}\right)_z=1 /\left(\frac{\partial x}{\partial y}\right)_z\)
(2) \(\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y=-1\)
(1) \(\left(\frac{\partial y}{\partial x}\right)_z=1 /\left(\frac{\partial x}{\partial y}\right)_z\)
\(\left(\frac{\partial y}{\partial x}\right)_z\)は\(z\)を固定したもとでの偏導関数。よって\(y\)は\(x\)だけの関数であり、1変数の関数である。
逆関数の微分公式
\[
\frac{d x}{d y}=1 / \frac{d y}{d x} \quad\left(\frac{d y}{d x} \neq 0 \text { のとき }\right)
\]
より
\[
\left(\frac{\partial y}{\partial x}\right)_z=1 /\left(\frac{\partial x}{\partial y}\right)_z
\]
(2) \(\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y=-1\)
略解
\(z\) は \(x, y\) の関数だから,
\[
d z=\left(\frac{\partial z}{\partial x}\right)_y d x+\left(\frac{\partial z}{\partial y}\right)_x d y
\]
ところで, \((\partial x / \partial y)_z\) は, \(z\) を一定, すなわち, \(d z=0\) としたときの \(d x / d y\) であるから,
上の式の両辺を \(d y\) で割って,
\[
0=\left(\frac{\partial z}{\partial x}\right)_y\left(\frac{\partial x}{\partial y}\right)_z+\left(\frac{\partial z}{\partial y}\right)_x
\]
(1)により, \((\partial z / \partial y)_x=1 /(\partial y / \partial z)_x\) だから $\( \left(\frac{\partial z}{\partial x}\right)_y\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x+1=0 \)$
よって，証明する式を得る.
なぜこうではないのか?：
定理より
\[\begin{split}
\left(\frac{\partial x}{\partial y}\right)_z = -\frac{F_y(x, y, z)}{F_x(x, y, z)}\\
\left(\frac{\partial y}{\partial z}\right)_x = -\frac{F_z(x, y, z)}{F_y(x, y, z)}\\
\left(\frac{\partial z}{\partial x}\right)_y = -\frac{F_x(x, y, z)}{F_z(x, y, z)}\\
\end{split}\]
問題の積は
\[\begin{split}
\begin{aligned}
\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y
&=
-\frac{F_y(x, y, z)}{F_x(x, y, z)}
\times
-\frac{F_z(x, y, z)}{F_y(x, y, z)}
\times
-\frac{F_x(x, y, z)}{F_z(x, y, z)}
\\
&=
(-1)^3
\frac{F_x(x, y, z) F_y(x, y, z) F_z(x, y, z) }{F_x(x, y, z) F_y(x, y, z)F_z(x, y, z)}
\\
&= -1
\end{aligned}
\end{split}\]
[6]#
[6] \(P(x, y) d x+Q(x, y) d y\) が全微分であるならば,
\[
\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}
\]
であることを示せ.
ある関数\(f(x,y)\)が存在して、
\[
df = P(x, y) dx + Q(x, y) dy
\]
であるとする。
\[
P(x, y) = \frac{\partial f}{\partial x} = f_x
,\quad
Q(x, y) = \frac{\partial f}{\partial y} = f_y
\]
であるため、
\[
\frac{\partial P}{\partial y} = f_{xy}
,\quad
\frac{\partial Q}{\partial x} = f_{yx}
\]
を意味する。
定理より、
\[
f_{xy} = f_{yx}
\]
であるため
\[
\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}
\]
が成り立つ。
[7]#
[7] 関数 \(F(x, y)\) がパラメータ \(\lambda\) と定数 \(p\) に対して,
\[
F(\lambda x, \lambda y)=\lambda^p F(x, y)
\]
をみたすとき, \(p\) 次の同次関数(homogeneous function)であるという. このとき, 次式 を証明せよ。
\[
x \frac{\partial F}{\partial x}+y \frac{\partial F}{\partial y}=p F
\]
Eulerの等式というらしい（参考）。
Eulerの等式
\(F: \mathbb{R}^2 \to \mathbb{R}\)が\(p\)次同次関数とする。このとき\(F(x,y)\)について、
\[
x \frac{\partial F}{\partial x}+y \frac{\partial F}{\partial y}=p F
\]
が成立する。
\[
F(\lambda x, \lambda y) = \lambda^p F(x, y)
\]
の両辺を\(\lambda\)で微分すると、
左辺は、\(F(\lambda x, \lambda y) = F(u(\lambda), v(\lambda))\)とおくと、全微分の合成関数の定理より
\[
\frac{dF(u(\lambda), v(\lambda))}{d\lambda}
= \frac{\partial F}{\partial u} \frac{d u}{d \lambda}
+ \frac{\partial F}{\partial v} \frac{d v}{d \lambda}
= \frac{\partial F}{\partial u} x
+ \frac{\partial F}{\partial v} y
\]
右辺は
\[
\frac{ d \lambda^p F(x, y) }{d \lambda}
= p\lambda^{p-1} F(x,y)
\]
整理すると
\[
\frac{\partial F}{\partial u} x
+ \frac{\partial F}{\partial v} y
= p\lambda^{p-1} F(x,y)
\]
ここで\(\lambda=1\)とすると、\(u=x, v=y\)なので
\[
\frac{\partial F}{\partial x} x
+ \frac{\partial F}{\partial y} y
= p F(x,y)
\]
[8]#
[8] \(x=r \sin \theta \cos \phi, y=r \sin \theta \sin \phi, z=r \cos \theta\) のとき, 次式を証明せよ.
\[
\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2}
=\frac{\partial^2 f}{\partial r^2}+\frac{2}{r} \frac{\partial f}{\partial r}+\frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2}+\frac{1}{r^2} \cot \theta \frac{\partial f}{\partial \theta}+\frac{1}{r^2 \sin ^2 \theta} \frac{\partial^2 f}{\partial \phi^2}
\]
左辺は\(f(x,y,z)\)（直交座標系）だが右辺は\(f(r,\theta,\phi)\)（球面座標系）となっている。両者の関連を導く。
\(x\)は\(r,\theta,\phi\)の関数なので、
1階微分は
\[
\frac{\partial f}{\partial x}
=\frac{\partial f}{\partial r} \frac{\partial r}{\partial x}
+\frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x}
+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x}
\]
2階微分は
\[\begin{split}
\begin{aligned}
\frac{\partial^2 f}{\partial^2 x}
&=\frac{\partial }{\partial x}
\left( \frac{\partial f}{\partial x} \right)\\
&=\frac{\partial }{\partial x}
\left(\frac{\partial f}{\partial r} \frac{\partial r}{\partial x}
+\frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x}
+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x}\right)
\\
&= \frac{\partial}{\partial r}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial r}{\partial x}
+\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \theta}{\partial x}
+\frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \phi}{\partial x}
\quad (おそらく、高階微分の順序を入れ替えられるため)
\end{aligned}
\end{split}\]
\(r,\theta,\phi\)が
\[
r=\sqrt{x^2+y^2+z^2}, \quad \theta=\arctan(\sqrt{x^2+y^2} / z), \quad \phi=\arctan (y / x)
\]
であることを知っていればまだ求められそう
（\(r\)の導関数）
\[\begin{split}
\begin{aligned}
\frac{\partial r}{\partial x}
&= \frac{1}{2} (x^2+y^2+z^2)^{-1/2} 2x
=\frac{x}{r}
=\sin \theta \cos \phi
\\
\frac{\partial r}{\partial y}
&=\frac{y}{r}
= \sin \theta \sin \phi
\\
\frac{\partial r}{\partial z}
&=\frac{z}{r}
= \cos \theta
\end{aligned}
\end{split}\]
\(=\sin \theta \cos \phi\)とかはよくわからない
（\(\theta\)の導関数）
公式より
\(\frac{d}{d x} \arctan x = \frac{1}{1+x^2}\)
なので
\(u=\sqrt{x^2+y^2} / z\)とおくと
\[\begin{split}
\begin{aligned}
\frac{\partial \theta}{\partial x}
&= \frac{\partial \arctan(u)}{\partial u} \cdot \frac{\partial u}{\partial x}\\
&= \frac{1}{1 + u^2} \cdot \frac{1}{z} \frac{1}{2} (x^2+y^2)^{-1/2} 2x\\
&= \frac{1}{1 + \frac{x^2+y^2}{z^2}} \cdot \frac{x}{z \sqrt{x^2+y^2}}\\
&= \frac{1}{\frac{x^2+y^2 + z^2}{z^2}} \cdot \frac{x}{z \sqrt{x^2+y^2}}\\
&= \frac{z^2}{r^2} \cdot \frac{x}{z \sqrt{x^2+y^2}}\\
&= \frac{zx}{r^2 \sqrt{x^2+y^2}}\\
\end{aligned}
\end{split}\]
同様に
\[\begin{split}
\begin{aligned}
\frac{\partial \theta}{\partial y}
&= \frac{zy}{r^2 \sqrt{x^2+y^2}}\\
\end{aligned}
\end{split}\]
またzは
\[\begin{split}
\begin{aligned}
\frac{\partial \theta}{\partial z}
&= \frac{\partial \arctan(u)}{\partial u} \cdot \frac{\partial u}{\partial z}\\
&= \frac{1}{1 + u^2} \cdot -\frac{1}{z^2} \sqrt{x^2+y^2} \\
&= \frac{z^2}{r^2} \cdot -\frac{\sqrt{x^2+y^2}}{z^2}\\
&= -\frac{\sqrt{x^2+y^2}}{r^2}\\
\end{aligned}
\end{split}\]
まとめると
\[
\frac{\partial \theta}{\partial x} = \frac{zx}{r^2 \sqrt{x^2+y^2}}
,\quad
\frac{\partial \theta}{\partial y}= \frac{zy}{r^2 \sqrt{x^2+y^2}}
,\quad
\frac{\partial \theta}{\partial z}= -\frac{\sqrt{x^2+y^2}}{r^2}
\]
（\(\phi\)の導関数）
公式より
\(\frac{d}{d x} \arctan x = \frac{1}{1+x^2}\)
なので
\[\begin{split}
\begin{aligned}
\frac{\partial \phi}{\partial x}
&= \frac{\partial \arctan(y/x)}{\partial (y/x)} \cdot \frac{\partial (y/x)}{\partial x}\\
&= \frac{1}{1+\frac{y^2}{x^2}} (-\frac{y}{x^2})\\
&= -\frac{y}{x^2+y^2}\\
\\
\frac{\partial \theta}{\partial y}
&= \frac{\partial \arctan(y/x)}{\partial (y/x)} \cdot \frac{\partial (y/x)}{\partial y}\\
&= \frac{1}{1+\frac{y^2}{x^2}} \cdot \frac{1}{x}\\
&= \frac{1}{\frac{x^2 + y^2}{x^2}} \cdot \frac{1}{x}\\
&= \frac{x^2}{x^2 + y^2} \cdot \frac{1}{x}\\
&= \frac{x}{x^2 + y^2}\\
\\
\frac{\partial \theta}{\partial z}
&=0
\end{aligned}
\end{split}\]
（複雑すぎてキャパオーバー
関連知識
問題冒頭の
\[\begin{split}
\begin{cases}
x=r \sin \theta \cos \phi\\
y=r \sin \theta \sin \phi\\
z=r \cos \theta
\end{cases}
\end{split}\]
は球面座標から直交座標への変換のこと（球面座標系 - Wikipedia）
直交座標系（Cartesian coordinates system)におけるLaplacian operator（勾配の2階微分バージョン）
\[
\Delta f =\nabla^2 f =\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2}
\]
球面座標系（spherical coordinates system）におけるLaplacian
\[
\Delta f =
\frac{\partial^2 f}{\partial r^2}+\frac{2}{r} \frac{\partial f}{\partial r}+\frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2}+\frac{1}{r^2} \cot \theta \frac{\partial f}{\partial \theta}+\frac{1}{r^2 \sin ^2 \theta} \frac{\partial^2 f}{\partial \phi^2}
\]