What is the derivative of 1/8 (e^{4x}+e^{-4x}) ?

The derivative of 1/8 (e^{4x}+e^{-4x}) is (e^{-4x}(e^{8x}-1))/2

What is the first derivative of 1/8 (e^{4x}+e^{-4x}) ?

The first derivative of 1/8 (e^{4x}+e^{-4x}) is (e^{-4x}(e^{8x}-1))/2

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derivative of

\frac{1}{8} (e^{4 x} + e^{- 4 x})

\frac{e ^{- 4 x} ( e ^{8 x} - 1 )}{2}

Solution steps

\frac{d}{d x} (\frac{1}{8} (e^{4 x} + e^{- 4 x}))

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (e^{4 x} + e^{- 4 x})

Apply the Sum/Difference Rule:

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

= \frac{1}{8} (\frac{d}{d x} (e^{4 x}) + \frac{d}{d x} (e^{- 4 x}))

\frac{d}{d x} (e^{4 x}) = e^{4 x} \cdot 4

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

= \frac{1}{8} (e^{4 x} \cdot 4 - 4 e^{- 4 x})

Simplify

\frac{1}{8} (e^{4 x} \cdot 4 - 4 e^{- 4 x}) : \frac{e ^{- 4 x} ( e ^{8 x} - 1 )}{2}

= \frac{e ^{- 4 x} ( e ^{8 x} - 1 )}{2}

What is the derivative of 1/8 (e^{4x}+e^{-4x}) ?
The derivative of 1/8 (e^{4x}+e^{-4x}) is (e^{-4x}(e^{8x}-1))/2
What is the first derivative of 1/8 (e^{4x}+e^{-4x}) ?
The first derivative of 1/8 (e^{4x}+e^{-4x}) is (e^{-4x}(e^{8x}-1))/2