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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

n = 1 \sum \infty \frac{2 n}{7 n - 2}

\int \frac{1}{1 + cos ( x )} d x

x \to 0 lim (\frac{1}{x} + \frac{1}{x ^{2}})

n = 1 \sum \infty (2 n)!

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