What is the d/(dz)(i/(2i)ln((i+iz)/(i-iz))) ?

The d/(dz)(i/(2i)ln((i+iz)/(i-iz))) is -1/((z-1)(z+1))

What is the first d/(dz)(i/(2i)ln((i+iz)/(i-iz))) ?

The first d/(dz)(i/(2i)ln((i+iz)/(i-iz))) is -1/((z-1)(z+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

\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}})

Simplify

\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 )}

\frac{d}{d x} (csc^{4} (tan (e^{2 x})))

y = 2 x^{3} + 3 x^{2} - 12 x + 1

\frac{\partial}{\partial x} (ln (3 x - 2 y))

\int cos (3 - 5 x) d x

\frac{d}{d x} (x \cdot e^{y})

What is the d/(dz)(i/(2i)ln((i+iz)/(i-iz))) ?
The d/(dz)(i/(2i)ln((i+iz)/(i-iz))) is -1/((z-1)(z+1))
What is the first d/(dz)(i/(2i)ln((i+iz)/(i-iz))) ?
The first d/(dz)(i/(2i)ln((i+iz)/(i-iz))) is -1/((z-1)(z+1))