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

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

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

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

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

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

Solution steps

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

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} (3 x^{2} - 8 x)

\frac{d y}{d x}, 2 x y = 0

\frac{d}{d x} (\frac{π ^{2}}{x ^{2} + 4 x})

3 x y + x^{2} + y^{2}

y^{^{''}} + 3 y^{^{'}} = (4 t + 4) sin (4 t)

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