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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int z cos (z) d z

9 y^{^{''}} + 10 y = 0, y (π) = 1

\frac{d}{d x} (9 x^{\frac{7}{5}} - 5 x^{2} + 1 0^{4})

x \to 0 + lim (- x + 7)

\frac{d ^{2} y}{d x ^{2}} = e^{5 x^{2}}

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