What is the derivative of 1/18 (e^{9x}+e^{-9x}) ?

The derivative of 1/18 (e^{9x}+e^{-9x}) is (e^{-9x}(e^{18x}-1))/2

What is the first derivative of 1/18 (e^{9x}+e^{-9x}) ?

The first derivative of 1/18 (e^{9x}+e^{-9x}) is (e^{-9x}(e^{18x}-1))/2

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\frac{d}{d x} (\frac{1}{18} (e^{9 x} + e^{- 9 x}))

\frac{e ^{- 9 x} ( e ^{18 x} - 1 )}{2}

Solution steps

\frac{d}{d x} (\frac{1}{18} (e^{9 x} + e^{- 9 x}))

Take the constant out:

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

= \frac{1}{18} \frac{d}{d x} (e^{9 x} + e^{- 9 x})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{1}{18} (\frac{d}{d x} (e^{9 x}) + \frac{d}{d x} (e^{- 9 x}))

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

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

= \frac{1}{18} (e^{9 x} \cdot 9 - 9 e^{- 9 x})

Simplify

\frac{1}{18} (e^{9 x} \cdot 9 - 9 e^{- 9 x}) : \frac{e ^{- 9 x} ( e ^{18 x} - 1 )}{2}

= \frac{e ^{- 9 x} ( e ^{18 x} - 1 )}{2}

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

f (x) = x^{2} + 8, (- 2, 12)

\frac{d}{d y} (- \frac{y}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}})

\int (3 x^{2} - x) d x

\frac{d}{d x} (e^{3 - 2 x})

What is the derivative of 1/18 (e^{9x}+e^{-9x}) ?
The derivative of 1/18 (e^{9x}+e^{-9x}) is (e^{-9x}(e^{18x}-1))/2
What is the first derivative of 1/18 (e^{9x}+e^{-9x}) ?
The first derivative of 1/18 (e^{9x}+e^{-9x}) is (e^{-9x}(e^{18x}-1))/2