What is the derivative of 1/10 (e^{5x}+e^{-5x}) ?

The derivative of 1/10 (e^{5x}+e^{-5x}) is (e^{-5x}(e^{10x}-1))/2

What is the first derivative of 1/10 (e^{5x}+e^{-5x}) ?

The first derivative of 1/10 (e^{5x}+e^{-5x}) is (e^{-5x}(e^{10x}-1))/2

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derivative of

\frac{1}{10} (e^{5 x} + e^{- 5 x})

\frac{e ^{- 5 x} ( e ^{10 x} - 1 )}{2}

Solution steps

\frac{d}{d x} (\frac{1}{10} (e^{5 x} + e^{- 5 x}))

Take the constant out:

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

= \frac{1}{10} \frac{d}{d x} (e^{5 x} + e^{- 5 x})

Apply the Sum/Difference Rule:

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

= \frac{1}{10} (\frac{d}{d x} (e^{5 x}) + \frac{d}{d x} (e^{- 5 x}))

\frac{d}{d x} (e^{5 x}) = e^{5 x} \cdot 5

\frac{d}{d x} (e^{- 5 x}) = - 5 e^{- 5 x}

= \frac{1}{10} (e^{5 x} \cdot 5 - 5 e^{- 5 x})

Simplify

\frac{1}{10} (e^{5 x} \cdot 5 - 5 e^{- 5 x}) : \frac{e ^{- 5 x} ( e ^{10 x} - 1 )}{2}

= \frac{e ^{- 5 x} ( e ^{10 x} - 1 )}{2}

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What is the derivative of 1/10 (e^{5x}+e^{-5x}) ?
The derivative of 1/10 (e^{5x}+e^{-5x}) is (e^{-5x}(e^{10x}-1))/2
What is the first derivative of 1/10 (e^{5x}+e^{-5x}) ?
The first derivative of 1/10 (e^{5x}+e^{-5x}) is (e^{-5x}(e^{10x}-1))/2