What is expand-e^x((xe^{-3x})/3+(e^{-3x})/9) ?

The solution to expand-e^x((xe^{-3x})/3+(e^{-3x})/9) is -(e^{-2x}x)/3-(e^{-2x})/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand-e^x((xe^{-3x})/3+(e^{-3x})/9)

expand

- e^{x} (\frac{x e ^{- 3 x}}{3} + \frac{e ^{- 3 x}}{9})

- \frac{e ^{- 2 x} x}{3} - \frac{e ^{- 2 x}}{9}

Solution steps

- e^{x} (\frac{x e ^{- 3 x}}{3} + \frac{e ^{- 3 x}}{9})

= - e^{x} (\frac{e ^{- 3 x} x}{3} + \frac{e ^{- 3 x}}{9})

Apply the distributive law:

a (b + c) = ab + a c

a = - e^{x}, b = \frac{x e ^{- 3 x}}{3}, c = \frac{e ^{- 3 x}}{9}

= - e^{x} \frac{x e ^{- 3 x}}{3} + (- e^{x}) \frac{e ^{- 3 x}}{9}

Apply minus-plus rules

+ (- a) = - a

= - e^{x} \frac{e ^{- 3 x} x}{3} - e^{x} \frac{e ^{- 3 x}}{9}

Simplify

- e^{x} \frac{e ^{- 3 x} x}{3} - e^{x} \frac{e ^{- 3 x}}{9} : - \frac{e ^{- 2 x} x}{3} - \frac{e ^{- 2 x}}{9}

= - \frac{e ^{- 2 x} x}{3} - \frac{e ^{- 2 x}}{9}

(2 x - x)^{2}

cos (2 (t + 2 π))

(e^{x}) (tan (x) - x)

(7 - 1) f t

\frac{x ^{5} + 2 x ^{3} - x - 8}{( x - 1 ) ^{2}}

What is expand-e^x((xe^{-3x})/3+(e^{-3x})/9) ?
The solution to expand-e^x((xe^{-3x})/3+(e^{-3x})/9) is -(e^{-2x}x)/3-(e^{-2x})/9