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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (- \frac{y}{x ^{2}} + 4 x + \frac{1}{y})

x^{2} \frac{d y}{d x} - 2 x y = 5 y^{4}

\int_{- 2}^{2} (x + 2) 4 - x^{2} d x

y^{^{''}} + 12 y^{^{'}} + 40 y = 0

\int - 2 e^{- 2 x} d x

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