What is expand (e^t)(e^{-t}+te^{-t}) ?

The solution to expand (e^t)(e^{-t}+te^{-t}) is 1+t

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expand (e^t)(e^{-t}+te^{-t})

expand

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

1 + t

Solution steps

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

= (e^{t}) (e^{- t} + e^{- t} t)

Apply the distributive law:

a (b + c) = ab + a c

a = (e^{t}), b = e^{- t}, c = t e^{- t}

= (e^{t}) e^{- t} + (e^{t}) t e^{- t}

= e^{- t} (e^{t}) + e^{- t} t (e^{t})

Simplify

e^{- t} (e^{t}) + e^{- t} t (e^{t}) : 1 + t

= 1 + t

\frac{1}{( x + a ) ( x - a )}

(2, 3) y (- 2, - 3)

\frac{( 2 x - 12 )}{( x - 4 ) ( x - 5 )}

x \cdot (\frac{1}{x} + \frac{y}{x ^{2}} - 3 y^{2})

(1 + \frac{3}{3}) (1 - \frac{3}{3})

What is expand (e^t)(e^{-t}+te^{-t}) ?
The solution to expand (e^t)(e^{-t}+te^{-t}) is 1+t