What is e^{-2t}(e^{-t+1}\H(t-1))' ?

The solution to e^{-2t}(e^{-t+1}\H(t-1))' is ((e^{-t+1}\H(t-1))')/(e^{2t)}

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simplify e^{-2t}(e^{-t+1}\H(t-1))'

simplify

e^{- 2 t} (e^{- t + 1} H (t - 1))^{'}

\frac{( e ^{- t + 1} H ( t - 1 ) ) '}{e ^{2 t}}

Solution steps

e^{- 2 t} (e^{- t + 1} H (t - 1))^{'}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{2 t}} (e^{- t + 1} H (t - 1))^{^{'}}

Multiply fractions:

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

= \frac{1 ( e ^{- t + 1} H ( t - 1 ) ) ^{^{'}}}{e ^{2 t}}

Multiply:

1 (e^{- t + 1} H (t - 1))^{'} = (e^{- t + 1} H (t - 1))^{'}

= \frac{( e ^{- t + 1} H ( t - 1 ) ) ^{^{'}}}{e ^{2 t}}

e^{5 x} e^{5 x}

sin (t) (cos (2 t))^{2} (cos (3 t))^{2}

e^{- \frac{17}{3}} \cdot (- 24)

- 1

What is e^{-2t}(e^{-t+1}\H(t-1))' ?
The solution to e^{-2t}(e^{-t+1}\H(t-1))' is ((e^{-t+1}\H(t-1))')/(e^{2t)}