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

The solution to expand 1+((e^x-e^{-x})^2)/4 is 1+(e^{2x})/4-1/2+(e^{-2x})/4

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expand 1+((e^x-e^{-x})^2)/4

expand

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

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

Solution steps

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

(e^{x} - e^{- x})^{2} = e^{2 x} - 2 + e^{- 2 x}

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

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{e ^{2 x} - 2 + e ^{- 2 x}}{4} = \frac{e ^{2 x}}{4} - \frac{2}{4} + \frac{e ^{- 2 x}}{4}

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

Cancel

\frac{2}{4} : \frac{1}{2}

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

\frac{d}{d y} ((x) (x - y))

(t e^{t} + t e^{2 t})^{2}

- \frac{40 ( - 3 x ^{2} - 4 )}{( x ^{2} - 4 ) ^{3}}

\frac{d}{d x} ((1 - x^{2}) arctan (x)) - x

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

What is expand 1+((e^x-e^{-x})^2)/4 ?
The solution to expand 1+((e^x-e^{-x})^2)/4 is 1+(e^{2x})/4-1/2+(e^{-2x})/4