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

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

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

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

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derivative of

y = \frac{e ^{9 x} + e ^{- 9 x}}{x}

\frac{9 e ^{9 x} x - 9 e ^{- 9 x} x - e ^{9 x} - e ^{- 9 x}}{x ^{2}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{d x}{d x} = 1

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

Simplify

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

= \frac{9 e ^{9 x} x - 9 e ^{- 9 x} x - e ^{9 x} - e ^{- 9 x}}{x ^{2}}

\int tan^{2} (4 x) d x

\int x x d x

\int_{- \infty}^{3} e^{0.0001 t} d t

\frac{\partial}{\partial y} (\frac{x}{y})

\frac{d}{d x} (e^{a x^{8}})

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