What is expand (1-e^{(-2v)})u(v) ?

The solution to expand (1-e^{(-2v)})u(v) is vu-e^{-2v}vu

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expand (1-e^{(-2v)})u(v)

expand

(1 - e^{(- 2 v)}) u (v)

vu - e^{- 2 v} vu

Solution steps

(1 - e^{(- 2 v)}) u (v)

= (v) u (1 - e^{(- 2 v)})

Apply the distributive law:

a (b - c) = ab - a c

a = u (v), b = 1, c = e^{(- 2 v)}

= u (v) \cdot 1 - u (v) e^{(- 2 v)}

= 1 \cdot (v) u - e^{(- 2 v)} (v) u

Simplify

1 \cdot (v) u - e^{(- 2 v)} (v) u : vu - e^{- 2 v} vu

= vu - e^{- 2 v} vu

\frac{3 x + 2}{2 x ( x + 1 )}

\frac{x - 7}{x ( x - 8 )}

\frac{5 e ^{- 2 t}}{s ( s ^{2} + 1 )}

\frac{( 7 )}{( 3 ( x ^{(2)} + 2 ) )}

k^{0}

What is expand (1-e^{(-2v)})u(v) ?
The solution to expand (1-e^{(-2v)})u(v) is vu-e^{-2v}vu