What is the derivative of 1/6 (e^{3x}+e^{-3x}) ?

The derivative of 1/6 (e^{3x}+e^{-3x}) is (e^{-3x}(e^{6x}-1))/2

What is the first derivative of 1/6 (e^{3x}+e^{-3x}) ?

The first derivative of 1/6 (e^{3x}+e^{-3x}) is (e^{-3x}(e^{6x}-1))/2

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\frac{d}{d x} (\frac{1}{6} (e^{3 x} + e^{- 3 x}))

\frac{e ^{- 3 x} ( e ^{6 x} - 1 )}{2}

Solution steps

\frac{d}{d x} (\frac{1}{6} (e^{3 x} + e^{- 3 x}))

Take the constant out:

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

= \frac{1}{6} \frac{d}{d x} (e^{3 x} + e^{- 3 x})

Apply the Sum/Difference Rule:

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

= \frac{1}{6} (\frac{d}{d x} (e^{3 x}) + \frac{d}{d x} (e^{- 3 x}))

\frac{d}{d x} (e^{3 x}) = e^{3 x} \cdot 3

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

= \frac{1}{6} (e^{3 x} \cdot 3 - 3 e^{- 3 x})

Simplify

\frac{1}{6} (e^{3 x} \cdot 3 - 3 e^{- 3 x}) : \frac{e ^{- 3 x} ( e ^{6 x} - 1 )}{2}

= \frac{e ^{- 3 x} ( e ^{6 x} - 1 )}{2}

What is the derivative of 1/6 (e^{3x}+e^{-3x}) ?
The derivative of 1/6 (e^{3x}+e^{-3x}) is (e^{-3x}(e^{6x}-1))/2
What is the first derivative of 1/6 (e^{3x}+e^{-3x}) ?
The first derivative of 1/6 (e^{3x}+e^{-3x}) is (e^{-3x}(e^{6x}-1))/2