What is the derivative of y=ln((e^x+1)/(e^x-1)) ?

The derivative of y=ln((e^x+1)/(e^x-1)) is -(2e^x)/((e^x+1)(e^x-1))

What is the first derivative of y=ln((e^x+1)/(e^x-1)) ?

The first derivative of y=ln((e^x+1)/(e^x-1)) is -(2e^x)/((e^x+1)(e^x-1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = ln (\frac{e ^{x} + 1}{e ^{x} - 1})

- \frac{2 e ^{x}}{( e ^{x} + 1 ) ( e ^{x} - 1 )}

Solution steps

\frac{d}{d x} (ln (\frac{e ^{x} + 1}{e ^{x} - 1}))

Apply the chain rule:

\frac{1}{\frac{e ^{x} + 1}{e ^{x} - 1}} \frac{d}{d x} (\frac{e ^{x} + 1}{e ^{x} - 1})

= \frac{1}{\frac{e ^{x} + 1}{e ^{x} - 1}} \frac{d}{d x} (\frac{e ^{x} + 1}{e ^{x} - 1})

\frac{d}{d x} (\frac{e ^{x} + 1}{e ^{x} - 1}) = - \frac{2 e ^{x}}{( e ^{x} - 1 ) ^{2}}

= \frac{1}{\frac{e ^{x} + 1}{e ^{x} - 1}} (- \frac{2 e ^{x}}{( e ^{x} - 1 ) ^{2}})

Simplify

\frac{1}{\frac{e ^{x} + 1}{e ^{x} - 1}} (- \frac{2 e ^{x}}{( e ^{x} - 1 ) ^{2}}) : - \frac{2 e ^{x}}{( e ^{x} + 1 ) ( e ^{x} - 1 )}

= - \frac{2 e ^{x}}{( e ^{x} + 1 ) ( e ^{x} - 1 )}

h (v) = 4 v^{2} - 2

f (x) = 9 x^{2} + 7 x - 6, at x = - \frac{7}{18}

\frac{d}{d x} (1 - ∣ x ∣)

\int e^{x} + cos (x) d x

\int \frac{5 x + 3}{7 x - 1} d x

What is the derivative of y=ln((e^x+1)/(e^x-1)) ?
The derivative of y=ln((e^x+1)/(e^x-1)) is -(2e^x)/((e^x+1)(e^x-1))
What is the first derivative of y=ln((e^x+1)/(e^x-1)) ?
The first derivative of y=ln((e^x+1)/(e^x-1)) is -(2e^x)/((e^x+1)(e^x-1))