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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (e^{2 x}) = e^{2 x} \cdot 2

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

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

Simplify

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

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

\frac{d}{d x} (\frac{x ^{3} - x ^{2} + 1}{5})

\int \frac{2}{1 + e ^{2 x}} d x

(cos^{2} (x) - 2 sin (x))^{^{'}}

\int \frac{sec ^{2} ( x )}{1 + 2 tan ( x )} d x

\int \frac{( 3 x ^{2} - 2 )}{x ( x + 1 ) ^{3}} d x

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