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

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

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

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

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derivative of

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

- e^{- x}

Solution steps

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

Simplify

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

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

Apply the chain rule:

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

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

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

= e^{- x} (- 1)

Simplify

= - e^{- x}

\int - x^{\frac{5}{2}} d x

\frac{d y}{d x} - 2 x y^{2} = 0

\int ln ((6 x)^{2}) d x

\int \frac{1}{sin ^{2} ( x )} d x

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0

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