What is the d/(dz)(ln(sqrt((b^2-z^2)/(b^2+z^2)))) ?

The d/(dz)(ln(sqrt((b^2-z^2)/(b^2+z^2)))) is -(2b^2z)/((z^2+b^2)(-z^2+b^2))

What is the first d/(dz)(ln(sqrt((b^2-z^2)/(b^2+z^2)))) ?

The first d/(dz)(ln(sqrt((b^2-z^2)/(b^2+z^2)))) is -(2b^2z)/((z^2+b^2)(-z^2+b^2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

- \frac{2 b ^{2} z}{( z ^{2} + b ^{2} ) ( - z ^{2} + b ^{2} )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d z} (\frac{b ^{2} - z ^{2}}{b ^{2} + z ^{2}}) = - \frac{2 b ^{2} z}{- z ^{2} + b ^{2} ( z ^{2} + b ^{2} ) ^{\frac{3}{2}}}

= \frac{1}{\frac{b ^{2} - z ^{2}}{b ^{2} + z ^{2}}} (- \frac{2 b ^{2} z}{- z ^{2} + b ^{2} ( z ^{2} + b ^{2} ) ^{\frac{3}{2}}})

Simplify

\frac{1}{\frac{b ^{2} - z ^{2}}{b ^{2} + z ^{2}}} (- \frac{2 b ^{2} z}{- z ^{2} + b ^{2} ( z ^{2} + b ^{2} ) ^{\frac{3}{2}}}) : - \frac{2 b ^{2} z}{( z ^{2} + b ^{2} ) ( - z ^{2} + b ^{2} )}

= - \frac{2 b ^{2} z}{( z ^{2} + b ^{2} ) ( - z ^{2} + b ^{2} )}

What is the d/(dz)(ln(sqrt((b^2-z^2)/(b^2+z^2)))) ?
The d/(dz)(ln(sqrt((b^2-z^2)/(b^2+z^2)))) is -(2b^2z)/((z^2+b^2)(-z^2+b^2))
What is the first d/(dz)(ln(sqrt((b^2-z^2)/(b^2+z^2)))) ?
The first d/(dz)(ln(sqrt((b^2-z^2)/(b^2+z^2)))) is -(2b^2z)/((z^2+b^2)(-z^2+b^2))