間違いしかありません.コメントにてご指摘いただければ幸いです(気が付いた点を特に断りなく頻繁に書き直していますのでご注意ください).

log(sin(πz))の微分

\(\log{\left(\sin{\left(\pi z\right)}\right)}\)の微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\log{\left(\sin{\left(\pi z\right)}\right)} &=&\frac{\mathrm{d}}{\mathrm{d}f}\log{\left(\sin{\left(f\right)}\right)}\frac{\mathrm{d}f}{\mathrm{d}z} \;\ldots\;f=\pi z,\;f\in\mathbb{C} \\&=&\frac{\mathrm{d}}{\mathrm{d}g}\log{\left(g\right)}\frac{\mathrm{d}g}{\mathrm{d}f}\frac{\mathrm{d}f}{\mathrm{d}z} \;\ldots\;g=\sin{\left(f\right)},\;g\in\mathbb{C} \\&=&\frac{1}{g}\;\cos{\left(f\right)}\;\pi \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/logz.html}{\frac{\mathrm{d}}{\mathrm{d}g}\log{\left(g\right)}=\frac{1}{g}} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/sinz.html}{\frac{\mathrm{d}}{\mathrm{d}f}\sin{\left(f\right)}=\cos{\left(f\right)}} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/az.html}{\frac{\mathrm{d}}{\mathrm{d}z}\pi z=\pi} \\&=&\frac{1}{\sin\left(f\right)}\;\cos{\left(f\right)}\;\pi \\&=&\frac{1}{\sin\left(\pi z\right)}\;\cos{\left(\pi z\right)}\;\pi \\&=&\pi\frac{\cos{\left(\pi z\right)}}{\sin\left(\pi z\right)} \\&=&\pi\frac{1}{\tan\left(\pi z\right)} \;\ldots\;\tan\left(\pi z\right)=\frac{\sin\left(\pi z\right)}{\cos{\left(\pi z\right)}} \\&=&\pi\cot{\left(\pi z\right)} \end{eqnarray}$$

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