d/(dz)(i/(2i)ln((i+iz)/(i-iz)))

解

\frac{d}{d z} (\frac{i}{2 i} ln (\frac{i + i z}{i - i z}))

- \frac{1}{( z - 1 ) ( z + 1 )}

解答ステップ

\frac{d}{d z} (\frac{i}{2 i} ln (\frac{i + i z}{i - i z}))

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{i}{2 i} \frac{d}{d z} (ln (\frac{i + i z}{i - i z}))

連鎖法則を適用:

\frac{1}{\frac{i + i z}{i - i z}} \frac{d}{d z} (\frac{i + i z}{i - i z})

= \frac{1}{\frac{i + i z}{i - i z}} \frac{d}{d z} (\frac{i + i z}{i - i z})

\frac{d}{d z} (\frac{i + i z}{i - i z}) = - \frac{2}{- 1 + 2 z - z ^{2}}

= \frac{i}{2 i} \cdot \frac{1}{\frac{i + i z}{i - i z}} (- \frac{2}{- 1 + 2 z - z ^{2}})

簡素化

\frac{i}{2 i} \cdot \frac{1}{\frac{i + i z}{i - i z}} (- \frac{2}{- 1 + 2 z - z ^{2}}) : - \frac{1}{( z - 1 ) ( z + 1 )}

= - \frac{1}{( z - 1 ) ( z + 1 )}