What is the derivative of 4/(e^x+e^{-x)} ?

The derivative of 4/(e^x+e^{-x)} is -(4(e^x-e^{-x}))/((e^x+e^{-x))^2}

What is the first derivative of 4/(e^x+e^{-x)} ?

The first derivative of 4/(e^x+e^{-x)} is -(4(e^x-e^{-x}))/((e^x+e^{-x))^2}

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derivative of

\frac{4}{e ^{x} + e ^{- x}}

- \frac{4 ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4}{e ^{x} + e ^{- x}})

Take the constant out:

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

= 4 \frac{d}{d x} (\frac{1}{e ^{x} + e ^{- x}})

Apply exponent rule:

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

= 4 \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}

= 4 (- \frac{1}{( e ^{x} + e ^{- x} ) ^{2}} (e^{x} - e^{- x}))

Simplify

4 (- \frac{1}{( e ^{x} + e ^{- x} ) ^{2}} (e^{x} - e^{- x})) : - \frac{4 ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}}

= - \frac{4 ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}}

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0

\frac{x}{( x ^{2} + 1 ) ( x ^{2} + 9 )}

\int t (t^{2} + 1)^{8} d t

\int_{1}^{e^{3}} \frac{ln ^{3} ( x ^{2} )}{x} d x

What is the derivative of 4/(e^x+e^{-x)} ?
The derivative of 4/(e^x+e^{-x)} is -(4(e^x-e^{-x}))/((e^x+e^{-x))^2}
What is the first derivative of 4/(e^x+e^{-x)} ?
The first derivative of 4/(e^x+e^{-x)} is -(4(e^x-e^{-x}))/((e^x+e^{-x))^2}