tanxと1/tan xの微分公式のいろいろな証明

$$\begin{aligned} &(\tan x)'\\ &=\left(\dfrac{\sin x}{\cos x}\right)'\\ &=\dfrac{(\sin x)'\cos x-(\cos x)'\sin x}{\cos^2 x}\\ &=\dfrac{\cos^2x+\sin^2x}{\cos^2 x}\\ &=\dfrac{1}{\cos^2x} \end{aligned}$$

$$\begin{aligned} &(\tan x)'\\ &=\left(\sin x \cdot \dfrac{1}{\cos x}\right)'\\ &= (\sin x)' \dfrac{1}{\cos x} + \sin x \left( \dfrac{1}{\cos x} \right)' \\ &= \dfrac{\cos x}{\cos x} - \sin x \dfrac{-\sin x}{\cos^2 x} \\ &= 1 + \tan^2 x\\ &=\dfrac{1}{\cos^2x} \end{aligned}$$

$$\begin{aligned} &(\tan x)'\\ &=\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h} \end{aligned}$$

ここで加法定理を使うと上式は，

$$\lim_{h\to 0}\dfrac{1}{h}\left(\dfrac{\tan x+\tan h}{1-\tan x\tan h}-\tan x\right)$$

となる。さらに通分して変形していく：

$$\begin{aligned} &\lim_{h\to 0}\dfrac{1}{h}\left(\dfrac{\tan h+\tan^2 x\tan h}{1-\tan x\tan h}\right)\\ &=\lim_{h\to 0}\dfrac{\tan h}{h}\cdot\dfrac{1+\tan^2 x}{1-\tan x\tan h}\\ &=\lim_{h\to 0}\dfrac{\sin h}{h}\cdot\dfrac{1}{\cos h}\cdot\dfrac{1}{1-\tan x\tan h}\cdot\dfrac{1}{\cos^2 x}\\ &=1\cdot 1\cdot \dfrac{1}{1-0}\cdot \dfrac{1}{\cos^2 x}\\ &=\dfrac{1}{\cos^2x} \end{aligned}$$

$$\begin{aligned} \left(\dfrac{1}{\tan x}\right)' &=\left(\dfrac{\cos x}{\sin x}\right)'\\ &=\dfrac{(\cos x)'\sin x-(\sin x)'\cos x}{\sin^2 x}\\ &=\dfrac{-\sin^2x-\cos^2x}{\sin^2 x}\\ &=-\dfrac{1}{\sin^2x} \end{aligned}$$

逆数の微分公式： $\left(\dfrac{1}{f(x)}\right)'=-\dfrac{f'(x)}{f(x)^2}$ より，

$$\begin{aligned} \left(\dfrac{1}{\tan x}\right)' &=-\dfrac{(\tan x)'}{\tan^2x}\\ &=-\dfrac{1}{\tan^2 x\cos^2 x}\\ &=-\dfrac{1}{\sin^2 x} \end{aligned}$$

$$\begin{aligned} \dfrac{1}{\tan x} &=\dfrac{\cos x}{\sin x}\\ &=\dfrac{\sin (x+\frac{\pi}{2})}{-\cos(x+\frac{\pi}{2})}\\ &=-\tan (x+\frac{\pi}{2}) \end{aligned}$$

より，

$$\begin{aligned} \left(\dfrac{1}{\tan x}\right)' &=-\dfrac{1}{\cos^2(x+\frac{\pi}{2})}\\ &=-\dfrac{1}{\sin^2x} \end{aligned}$$