What is the derivative of ln((e^x+e^{-x}/2)) ?

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

What is the first derivative of ln((e^x+e^{-x}/2)) ?

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

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Popular >

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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\int 1. 8^{0.045 \cdot x} - 1 d x

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\int \frac{1}{x ^{3} x ^{2} - 36} d x

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