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

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

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

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

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derivative of

\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{d y}{d x} = 2 x y^{2} + 128 x

x \to - 6 lim (12 x + 1)

\frac{d y}{d x} = \frac{1}{2} (y - 1)^{2}, y (0) = 2

\int \frac{x ^{3}}{x ^{2} + 4} d x

\frac{d}{d x} (- \frac{2}{( x - 1 ) ^{3}})

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