What is the derivative of 9/(e^x-e^{-x}) ?

The derivative of 9/(e^x-e^{-x}) is -(9(e^x+e^{-x}))/((e^x-e^{-x))^2}

What is the first derivative of 9/(e^x-e^{-x}) ?

The first derivative of 9/(e^x-e^{-x}) is -(9(e^x+e^{-x}))/((e^x-e^{-x))^2}

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\frac{d}{d x} (\frac{9}{e ^{x} - e ^{- x}})

- \frac{9 ( e ^{x} + e ^{- x} )}{( e ^{x} - e ^{- x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{9}{e ^{x} - e ^{- x}})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{1}{e ^{x} - e ^{- x}})

Apply exponent rule:

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

= 9 \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}

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

Simplify

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

= - \frac{9 ( e ^{x} + e ^{- x} )}{( e ^{x} - e ^{- x} ) ^{2}}

x^{2} + 3 y \frac{d y}{d x} = 0

\int_{0}^{16} \frac{1}{x} d x

y^{^{'}} = ry

y^{^{''}} = - \frac{2}{t ^{2}}

y = x^{2} + 5 x + 9, at x = 4

What is the derivative of 9/(e^x-e^{-x}) ?
The derivative of 9/(e^x-e^{-x}) is -(9(e^x+e^{-x}))/((e^x-e^{-x))^2}
What is the first derivative of 9/(e^x-e^{-x}) ?
The first derivative of 9/(e^x-e^{-x}) is -(9(e^x+e^{-x}))/((e^x-e^{-x))^2}