What is the d/(dy)((e^y+e^{-y})/2) ?

The d/(dy)((e^y+e^{-y})/2) is 1/2 (e^y-e^{-y})

What is the first d/(dy)((e^y+e^{-y})/2) ?

The first d/(dy)((e^y+e^{-y})/2) is 1/2 (e^y-e^{-y})

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

\frac{1}{2} (e^{y} - e^{- y})

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d y} (e^{y}) = e^{y}

\frac{d}{d y} (e^{- y}) = - e^{- y}

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

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

x \to 4 lim (\frac{x + 4}{x + 7})

\frac{d}{d x} (3 x^{5 x})

t a y l or arctan (2 x)

y^{^{'}} = x y^{3}

What is the d/(dy)((e^y+e^{-y})/2) ?
The d/(dy)((e^y+e^{-y})/2) is 1/2 (e^y-e^{-y})
What is the first d/(dy)((e^y+e^{-y})/2) ?
The first d/(dy)((e^y+e^{-y})/2) is 1/2 (e^y-e^{-y})