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