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) $$\begin{eqnarray} \\y&=&\tan{\left(x\right)} \\\frac{\mathrm{d}}{\mathrm{d}x}\tan{\left(x\right)} &=&\lim_{h\rightarrow0}\frac{\tan{\left(x+h\right)}-\tan{\left(x\right)}}{h} \\&=&\lim_{h\rightarrow0}\frac{1}{h}\left\{ \frac{\tan{\left(x\right)}+\tan{\left(h\right)}}{1-\tan{\left(x\right)}\tan{\left(h\right)}} -\tan{\left(x\right)} \right\} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(x\right)}+\tan{\left(h\right)} -\tan{\left(x\right)}\left(1-\tan{\left(x\right)}\tan{\left(h\right)}\right) }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(x\right)}+\tan{\left(h\right)} -\tan{\left(x\right)}+\tan^2{\left(x\right)}\tan{\left(h\right)} }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(h\right)}+\tan^2{\left(x\right)}\tan{\left(h\right)} }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(h\right)}\left(1+\tan^2{\left(x\right)}\right) }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{\tan{\left(h\right)}}{h} \frac{1+\tan^2{\left(x\right)}}{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&1\cdot\frac{1+\tan^2{\left(x\right)}}{1-\tan{\left(x\right)}\cdot0} \\&&\;\ldots\;\lim_{h\rightarrow0}\frac{\tan{\left(h\right)}}{h} =\lim_{h\rightarrow0}\frac{1}{h}\frac{\sin{\left(h\right)}}{\cos{\left(h\right)}} =\lim_{h\rightarrow0}\frac{\sin{\left(h\right)}}{h}\frac{1}{\cos{\left(h\right)}} =1\cdot1 =1 \\&&\;\ldots\;\lim_{h\rightarrow0}\frac{\sin{\left(h\right)}}{h}=1 \\&&\;\ldots\;\lim_{h\rightarrow0}\cos{\left(h\right)}=1 \\&=&1+\tan^2{\left(x\right)} \end{eqnarray}$$ $$\begin{eqnarray} 1+\tan^2{\left(x\right)}&=&1+\left\{\frac{\sin{\left(x\right)}}{\cos{\left(x\right)}}\right\}^2 \\&=&\frac{\cos^2{\left(x\right)}+\sin^2{\left(x\right)}}{\cos^2{\left(x\right)}} \\&=&\frac{1}{ \cos^2{\left(x\right)} } \;\cdots\;\cos^2{\left(x\right)}+\sin^2{\left(x\right)}=1 \\&=&\sec^{2}{\left(x\right)} \end{eqnarray}$$