逆凾数の微分 (微分可能な凾数の逆凾数の微分)

逆凾数の微分 (微分可能な凾数の逆凾数の微分)

凾数，逆凾数における独立変数の差と従属変数の差の関係

$$\begin{eqnarray} y&=&f\left(x\right)\;\ldots\;微分可能な凾数 \\x&=&f^{-1}\left(y\right)\;\ldots\;f(x)の逆凾数 \end{eqnarray}$$ $$\begin{eqnarray} k&=&f\left(x+h\right)-f\left(x\right) \\&&\;\ldots\;凾数において独立変数の差hがある際の従属変数の差をkとする \\f\left(x\right)+k&=&y+k \\&=&f\left(x+h\right) \\x+h&=&f^{-1}\left(y+k\right) \end{eqnarray}$$ $$\begin{eqnarray} x+h&=&f^{-1}\left(y+k\right) \\h&=&f^{-1}\left(y+k\right)-x \\&=&f^{-1}\left(y+k\right)-f^{-1}\left(y\right) \\&&\;\ldots\;逆凾数において独立変数の差kがある際の従属変数の差がhである \end{eqnarray}$$ $$\begin{eqnarray} \lim_{k\rightarrow0}f^{-1}\left(y+k\right)&=&f^{-1}\left(y\right) \\h&=&\lim_{k\rightarrow0}\left(f^{-1}\left(y+k\right)-f^{-1}\left(y\right)\right) \\&=&\left(\lim_{k\rightarrow0}f^{-1}\left(y+k\right)\right)-f^{-1}\left(y\right) \\&=&f^{-1}\left(y\right)-f^{-1}\left(y\right) \\&=&0 \\&&\;\ldots\;kを0に近づけることはhを0に近づけることと等しい \end{eqnarray}$$ $$\begin{eqnarray} k&=&f\left(x+h\right)-f\left(x\right) \\\lim_{h\rightarrow0}f\left(x+h\right)&=&f\left(x\right) \\k&=&\lim_{h\rightarrow0}\left(f\left(x+h\right)-f\left(x\right)\right) \\&=&\left(\lim_{h\rightarrow0}f\left(x+h\right)\right)-f\left(x\right) \\&=&f\left(x\right)-f\left(x\right) \\&=&0 \\&&\;\ldots\;hを0に近づけることはkを0に近づけることと等しい \end{eqnarray}$$ $$\begin{eqnarray} h\rightarrow0 \iff k\rightarrow0 \\\lim_{k\rightarrow0}\frac{h}{k}=\lim_{h\rightarrow0}\frac{h}{k} \end{eqnarray}$$

逆凾数の微分

$$\begin{eqnarray} \frac{\mathrm{d}x}{\mathrm{d}y}&=&\lim_{k\rightarrow0}\frac{f^{-1}\left(y+k\right)-f^{-1}\left(y\right)}{k} \\&=&\lim_{k\rightarrow0}\frac{x+h-x}{k} \\&=&\lim_{k\rightarrow0}\frac{h}{k} \\&=&\lim_{h\rightarrow0}\frac{h}{k} \\&=&\lim_{h\rightarrow0}\frac{1}{\frac{k}{h}} \\&=&\frac{1}{\lim_{h\rightarrow0}\frac{k}{h}} \\&=&\frac{1}{\lim_{h\rightarrow0}\frac{f\left(x+h\right)-f\left(x\right)}{h}} \\&=&\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}} \end{eqnarray}$$

tan(x)の微分 2(微分の定義からの場合)

\(\tan{\left(x\right)}\)の微分 2

(sin, cosの微分を既知とした場合) $$\begin{eqnarray} \\y&=&\tan{\left(x\right)} \\\frac{\mathrm{d}}{\mathrm{d}x}\tan{\left(x\right)} &=&\lim_{h\rightarrow0}\frac{\tan{\left(x+h\right)}-\tan{\left(x\right)}}{h} \\&=&\lim_{h\rightarrow0}\frac{1}{h}\left\{ \frac{\tan{\left(x\right)}+\tan{\left(h\right)}}{1-\tan{\left(x\right)}\tan{\left(h\right)}} -\tan{\left(x\right)} \right\} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(x\right)}+\tan{\left(h\right)} -\tan{\left(x\right)}\left(1-\tan{\left(x\right)}\tan{\left(h\right)}\right) }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(x\right)}+\tan{\left(h\right)} -\tan{\left(x\right)}+\tan^2{\left(x\right)}\tan{\left(h\right)} }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(h\right)}+\tan^2{\left(x\right)}\tan{\left(h\right)} }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(h\right)}\left(1+\tan^2{\left(x\right)}\right) }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{\tan{\left(h\right)}}{h} \frac{1+\tan^2{\left(x\right)}}{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&1\cdot\frac{1+\tan^2{\left(x\right)}}{1-\tan{\left(x\right)}\cdot0} \\&&\;\ldots\;\lim_{h\rightarrow0}\frac{\tan{\left(h\right)}}{h} =\lim_{h\rightarrow0}\frac{1}{h}\frac{\sin{\left(h\right)}}{\cos{\left(h\right)}} =\lim_{h\rightarrow0}\frac{\sin{\left(h\right)}}{h}\frac{1}{\cos{\left(h\right)}} =1\cdot1 =1 \\&&\;\ldots\;\lim_{h\rightarrow0}\frac{\sin{\left(h\right)}}{h}=1 \\&&\;\ldots\;\lim_{h\rightarrow0}\cos{\left(h\right)}=1 \\&=&1+\tan^2{\left(x\right)} \end{eqnarray}$$ $$\begin{eqnarray} 1+\tan^2{\left(x\right)}&=&1+\left\{\frac{\sin{\left(x\right)}}{\cos{\left(x\right)}}\right\}^2 \\&=&\frac{\cos^2{\left(x\right)}+\sin^2{\left(x\right)}}{\cos^2{\left(x\right)}} \\&=&\frac{1}{ \cos^2{\left(x\right)} } \;\cdots\;\cos^2{\left(x\right)}+\sin^2{\left(x\right)}=1 \\&=&\sec^{2}{\left(x\right)} \end{eqnarray}$$

tan(x)の微分 1 (sin, cosの微分を既知とした場合)

\(\tan{\left(x\right)}\)の微分 1

(微分の定義からの場合) $$\begin{eqnarray} y&=&\tan{\left(x\right)} \\\frac{\mathrm{d}}{\mathrm{d}x}\tan{\left(x\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\frac{\sin{\left(x\right)}}{\cos{\left(x\right)}} \\&=&\frac{\mathrm{d}}{\mathrm{d}x}\sin{\left(x\right)}\left\{\cos{\left(x\right)}\right\}^{-1} \\&=&\left\{\cos{\left(x\right)}\right\}^{-1}\frac{\mathrm{d}}{\mathrm{d}x}\sin{\left(x\right)} +\sin{\left(x\right)}\frac{\mathrm{d}}{\mathrm{d}x}\left\{\cos{\left(x\right)}\right\}^{-1} \\&=&\left\{\cos{\left(x\right)}\right\}^{-1}\left\{\cos\left(x\right)\right\} +\sin{\left(x\right)} \left[-\left\{\cos{\left(x\right)}\right\}^{-2}\right] \frac{\mathrm{d}}{\mathrm{d}x}\left\{\cos{\left(x\right)}\right\} \\&=&1+\frac{-\sin{\left(x\right)}}{\cos^2{\left(x\right)}} \left\{-\sin{\left(x\right)}\right\} \\&=&1+\frac{\sin^2{\left(x\right)}}{\cos^2{\left(x\right)}} \\&=&1+\left\{\frac{\sin{\left(x\right)}}{\cos{\left(x\right)}}\right\}^2 \\&=&1+\tan^2{\left(x\right)} \end{eqnarray}$$

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$$\begin{eqnarray} 1+\tan^2{\left(x\right)}&=&1+\left\{\frac{\sin{\left(x\right)}}{\cos{\left(x\right)}}\right\}^2 \\&=&\frac{\cos^2{\left(x\right)}+\sin^2{\left(x\right)}}{\cos^2{\left(x\right)}} \\&=&\frac{1}{ \cos^2{\left(x\right)} } \;\cdots\;\cos^2{\left(x\right)}+\sin^2{\left(x\right)}=1 \\&=&\sec^{2}{\left(x\right)} \end{eqnarray}$$