What is the derivative of 1/14 (e^{7x}+e^{-7x}) ?

The derivative of 1/14 (e^{7x}+e^{-7x}) is (e^{-7x}(e^{14x}-1))/2

What is the first derivative of 1/14 (e^{7x}+e^{-7x}) ?

The first derivative of 1/14 (e^{7x}+e^{-7x}) is (e^{-7x}(e^{14x}-1))/2

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\frac{d}{d x} (\frac{1}{14} (e^{7 x} + e^{- 7 x}))

\frac{e ^{- 7 x} ( e ^{14 x} - 1 )}{2}

Solution steps

\frac{d}{d x} (\frac{1}{14} (e^{7 x} + e^{- 7 x}))

Take the constant out:

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

= \frac{1}{14} \frac{d}{d x} (e^{7 x} + e^{- 7 x})

Apply the Sum/Difference Rule:

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

= \frac{1}{14} (\frac{d}{d x} (e^{7 x}) + \frac{d}{d x} (e^{- 7 x}))

\frac{d}{d x} (e^{7 x}) = e^{7 x} \cdot 7

\frac{d}{d x} (e^{- 7 x}) = - 7 e^{- 7 x}

= \frac{1}{14} (e^{7 x} \cdot 7 - 7 e^{- 7 x})

Simplify

\frac{1}{14} (e^{7 x} \cdot 7 - 7 e^{- 7 x}) : \frac{e ^{- 7 x} ( e ^{14 x} - 1 )}{2}

= \frac{e ^{- 7 x} ( e ^{14 x} - 1 )}{2}

\int_{0}^{1} π (tan (\frac{π}{4} y))^{2} d y

\frac{\partial}{\partial x} (\frac{5 x + 4 y}{3 x + 2 y})

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

\frac{\partial}{\partial x} (6 - x^{3} - y^{5})

\frac{d}{d x} (arcsec (7 x))

What is the derivative of 1/14 (e^{7x}+e^{-7x}) ?
The derivative of 1/14 (e^{7x}+e^{-7x}) is (e^{-7x}(e^{14x}-1))/2
What is the first derivative of 1/14 (e^{7x}+e^{-7x}) ?
The first derivative of 1/14 (e^{7x}+e^{-7x}) is (e^{-7x}(e^{14x}-1))/2