数学-微分公式の証明

# 微分公式の証明

[数学に戻る](eb2ee12608a969cea3e4)

## 目次

[(1)](#1) $ (x^n)' = n x^{n - 1} $
[(2)](#2) $ (\sin x)' = \cos x, (\sin ax)' = a \cos x $
[(3)](#3) $ (\cos x)' = - \sin x, (\cos ax)' = -a \sin x $
[(4)](#4) $ (\tan x)' = \frac{1}{\cos^2 x} $
[(5)](#5) $ (e^x)' = e^x $
[(6)](#6) $ (a^x)' = a^x \log a $
[(7)](#7) $ (\log x)' = \frac{1}{x} (x > 0)$
[(8)](#8) $ \\{ \log f(x) \\}' = \frac{f(x)'}{f(x)} (f(x) > 0)$

## 証明

### (1)

$ (x^n)' = n x^{n - 1} $

----

- 二項定理を用いた定番の証明

```math
\begin{array}{ll}
&微分の定義より、f(x) = x^n の導関数 f(x)' は\\
f(x)' &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\
&= \lim_{h \to 0} \frac{(x + h)^n - x^n}{h}\\
&この分子を二項定理を用いて展開すると\\
&= \lim_{h \to 0} \frac{(x^n + nhx^{n-1} + {}_n C_2 h^2 x^{n-2} + \cdots + h^n) - x^n}{h}\\
&= \lim_{h \to 0} \frac{nhx^{n-1} + {}_n C_2 h^2 x^{n-2} + \cdots + h^n}{h}\\
&= \lim_{h \to 0} nx^{n-1} + {}_n C_2 h x^{n-2} + \cdots + h^{n-1}\\ 
&= nx^{n-1}
\end{array}
```

- 数学的帰納法による証明

```math
\begin{array}{lrl}
n = 1 &のとき、&f(x) = x となり\\
&&f'(x) = \lim_{h \to 0} \frac{(x + h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1\\
&&nx^{n-1} = 1 \cdot x^{1-1} = 1\\
&&で成り立つ。\\
\\
n = k &で成り立つと&仮定すると、(x^k)' = kx^{k-1}\\
&(x^{k+1})' &= (x \cdot x^k)' \\
&&= (x)' \cdot x^k + x \cdot (x^k)'\\
&&= 1 \cdot x^k + x \cdot kx^{k-1}\\
&&= x^k + kx^k\\
&&= (k+1)x^k\\
&n = k+1 のときも成り立つ & \therefore nx^{n-1}
\end{array}
```

### (2)

$ (\sin x)' = \cos x $
$ (\sin ax)' = a \cos x $

----

```math
\begin{array}{ll}

f(x)' &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\\
&= \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}\\
&= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}\\
&= \lim_{h \to 0} \left( \cos x \cdot \frac{\sin h}{h} - \sin x \cdot \frac{1 - \cos h}{h} \right)\\
&= \lim_{h \to 0} \left( \cos x \cdot \frac{\sin h}{h} - \sin x \cdot h \cdot \frac{1 - \cos h}{h^2} \right)\\
&= \cos x \cdot 1 - \sin x \cdot 0 \cdot \frac{1}{2}\\
&= \cos x \\

\\

f(x)' &= (\sin ax)' \cdot (ax)' \because 合成関数の微分 \\
&= a \cos x \\

\end{array}
```


### (3)

$ (\cos x)' = - \sin x $
$ (\cos ax)' = -a \sin x $

----

```math
\begin{array}{ll}

f(x)' &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\\
&= \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}\\
&= \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}\\
&= \lim_{h \to 0} \left(- \sin x \cdot \frac{\sin h}{h} - \cos x \cdot \frac{1 - \cos h}{h} \right)\\
&= \lim_{h \to 0} \left(- \sin x \cdot \frac{\sin h}{h} - \cos x \cdot h \cdot \frac{1 - \cos h}{h^2} \right)\\
&= - \sin x \cdot 1 - \cos x \cdot 0 \cdot \frac{1}{2}\\
&= - \sin x\\
\\

f(x)' &= (\cos ax)' \cdot (ax)' \because 合成関数の微分\\
&= -a \sin x \\

\end{array}
```

### (4)

$ (\tan x)' = \frac{1}{\cos^2 x} $

----

```math
\begin{array}{ll}

(\tan x)' &= (\frac{\sin x}{\cos x})'\\
&= \frac{(\sin x)' \cdot \cos x - \sin x \cdot (\cos x)'}{\cos^2 x}\\
&= \frac{\cos x \cdot \cos x - \sin x \cdot (- \sin x)}{\cos^2 x}\\
&= \frac{\cos^2 x + \sin^2 x}{\cos^2 x}\\
&= \frac{1}{\cos^2 x}\\

\end{array}
```

### (5)

$ (e^x)' = e^x $

----

```math
\begin{array}{ll}

f(x)' &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\\
&= \lim_{h \to 0} \frac{e^{x+h} - e^x}{h}\\
&= \lim_{h \to 0} \frac{e^x(e^h - 1)}{h}\\
&= \lim_{h \to 0} e^x \cdot \frac{(e^h - 1)}{h}\\
&= e^x　\because 指数・対数関数の極限公式 \lim_{h \to 0} \frac{(e^h - 1)}{h} = 1\\

\end{array}
```

### (6)

$ (a^x)' = a^x \log a $

----

```math
\begin{array}{ll}

f(x)' &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\\
&= \lim_{h \to 0} \frac{a^{x+h} - a^x}{h}\\
&= \lim_{h \to 0} \frac{a^x(a^h - 1)}{h}\\
&= a^x \cdot \lim_{h \to 0} \frac{a^h - 1}{h}\\
&= a^x \cdot \lim_{h \to 0} \frac{e^{\log a^h} - 1}{\log a^h} \cdot \frac{\log a^h}{h}\\
& u = \log a^h とおくと、{h \to 0} のとき {u \to 0}\\
&= a^x \cdot \lim_{u \to 0} \frac{e^u - 1}{u} \cdot \frac{\log a^h}{h}\\
&= a^x \cdot 1 \cdot \frac{\log a^h}{h}\\
&= a^x \cdot \frac{h \cdot \log a}{h}\\
&= a^x \cdot \log a\\
\\
&a^h = e^{\log a^h}　\because 両辺の \log をとるとおなじになる
\end{array}
```

### (7)

$ (\log x)' = \frac{1}{x} (x > 0)$

----

```math
\begin{array}{ll}

f(x)' &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\\
&= \lim_{h \to 0} \frac{\log (x+h) - \log x}{h}\\
&= \lim_{h \to 0} \frac{\log \left( \frac{x+h}{x} \right)}{h}\\
&= \lim_{h \to 0} \frac{\log \left( 1 + \frac{h}{x} \right)}{h}\\
&= \lim_{h \to 0} \frac{1}{h} \cdot \log \left( 1 + \frac{h}{x} \right)\\
&= \lim_{h \to 0} \frac{1}{x} \cdot \frac{x}{h} \cdot \log \left( 1 + \frac{h}{x} \right)\\
&= \lim_{h \to 0} \frac{1}{x} \cdot \log \left( 1 + \frac{h}{x} \right)^{\frac{x}{h}}\\
& u = \frac{h}{x} とおくと、{h \to 0} のとき {u \to 0}\\
&= \lim_{u \to 0} \frac{1}{x} \cdot \log \left( 1 + u \right)^{\frac{1}{u}}　\because \lim_{u \to 0} \left( 1 + u \right)^{\frac{1}{u}} = e\\
&= \frac{1}{x} \cdot \log e　\because \log e = 1\\
&= \frac{1}{x}

\end{array}
```

### (8)

$ \\{ \log f(x) \\}' = \frac{f(x)'}{f(x)} (f(x) > 0)$

----

```math

y = \log f(x)、t = f(x) とおくと\\

\begin{array}{ll}

\{ \log f(x) \}' &= y'\\
&= \frac{dy}{dt} \cdot \frac{dt}{dx}\\
&= \frac{d(\log t)}{dt} \cdot \frac{d\{f(x)\}}{dx} 
&= \frac{1}{t} \cdot f'(x)\\
&= \frac{f(x)}{f'(x)}\\

\end{array}
```

## 微分の重要公式

### (1)

$ \\{ f(x) \cdot g(x) \\}' = f'(x) \cdot g(x) + f(x) \cdot g'(x) $

----

```math
\begin{array}{ll}

\{ f(x) \cdot g(x) \}' &= \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}\\
&= \lim_{h \to 0} \frac{f(x+h)g(x+h) + \{ - f(x)g(x+h) + f(x)g(x+h) \} - f(x)g(x)}{h}\\
&= \lim_{h \to 0} \frac{\{f(x+h) - f(x) \} \cdot g(x+h) + f(x) \cdot \{g(x+h) - g(x) \}}{h}\\
&= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \cdot g(x+h) + f(x) \cdot \frac{g(x+h) - g(x)}{h}\\
&= f'(x) \cdot g(x) + f(x) \cdot g'(x)\\

\end{array}
```

### (2)

$ \left( \frac{f(x)}{g(x)} \right)' = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{\\{g(x)\\}^2} $

----

```math
\begin{array}{ll}

\left( \frac{f(x)}{g(x)} \right)' &= \lim_{h \to 0} \frac{\frac{f(x+h)}{g(x+h)} -\frac{f(x)}{g(x)}}{h}\\
&= \lim_{h \to 0} \frac{f(x+h)g(x) - f(x)g(x+h)}{h \cdot g(x+h) \cdot g(x)}\\
&= \lim_{h \to 0} \frac{f(x+h)g(x) + \{ - f(x)g(x) + f(x)g(x) \} - f(x)g(x+h)}{h \cdot g(x+h) \cdot g(x)}\\
&= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \cdot \frac{g(x)}{g(x+h) \cdot g(x)} - \frac{f(x)}{g(x+h) \cdot g(x)} \cdot \frac{g(x+h) - g(x)}{h}\\
&= f'(x) \cdot \frac{1}{g(x)} - \frac{f(x)}{\{g(x)\}^2} \cdot g'(x)\\
&= \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{\{g(x)\}^2}\\

\end{array}
```

### (3)

$ \\{ f(g(x)) \\}' = f'(g(x)) \cdot g'(x) $

----

```math

\begin{array}{ll}

\{ f(g(x)) \}' &= \lim_{h \to 0} \frac{f(g(x+h)) - f(g(x))}{h}\\
& g(x+h) - g(x) = k とおくと g(x+h) = g(x) + k\\
& {h \to 0} のとき {k \to 0}\\
&= \lim_{h \to 0} \frac{f(g(x) + k) - f(g(x))}{h}\\
& また、g(x) = u とおく\\
&= \lim_{h \to 0} \frac{f(u + k) - f(u)}{h}\\
&= \lim_{h \to 0} \frac{f(u + k) - f(u)}{k} \cdot \frac{k}{h}\\
&= \lim_{h \to 0, k \to 0} \frac{f(u + k) - f(u)}{k} \cdot \frac{g(x+h) - g(x)}{h}\\
&= f'(u) \cdot g'(x)\\
&= f'(g(x)) \cdot g'(x)\\

\end{array}

```