What is expand e^{2(t-1)}cos(w)(t-1)\H(t-1) ?

The solution to expand e^{2(t-1)}cos(w)(t-1)\H(t-1) is e^{2t-2}t\H(t-1)cos(w)-e^{2t-2}\H(t-1)cos(w)

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expand e^{2(t-1)}cos(w)(t-1)\H(t-1)

expand

e^{2 (t - 1)} cos (w) (t - 1) H (t - 1)

e^{2 t - 2} t H (t - 1) cos (w) - e^{2 t - 2} H (t - 1) cos (w)

Solution steps

e^{2 (t - 1)} cos (w) (t - 1) H (t - 1)

= e^{2 (t - 1)} H (t - 1) cos (w) (t - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = e^{2 (t - 1)} cos (w) H (t - 1), b = t, c = 1

= e^{2 (t - 1)} cos (w) H (t - 1) t - e^{2 (t - 1)} cos (w) H (t - 1) \cdot 1

= e^{2 (t - 1)} t H (t - 1) cos (w) - 1 \cdot e^{2 (t - 1)} H (t - 1) cos (w)

Simplify

e^{2 (t - 1)} t H (t - 1) cos (w) - 1 \cdot e^{2 (t - 1)} H (t - 1) cos (w) : e^{2 t - 2} t H (t - 1) cos (w) - e^{2 t - 2} H (t - 1) cos (w)

= e^{2 t - 2} t H (t - 1) cos (w) - e^{2 t - 2} H (t - 1) cos (w)

\frac{- l ( - 2 l ( - 2 l ( - 2 l + 1 ) - 1 ) + 1 ) - 1}{8}

\frac{( x - 1 ) lo g _{2} ( x )}{1 + x}

\frac{3 ( 1 + 5 )}{2}

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

\frac{9 ( x ^{10} - 1 )}{5}

What is expand e^{2(t-1)}cos(w)(t-1)\H(t-1) ?
The solution to expand e^{2(t-1)}cos(w)(t-1)\H(t-1) is e^{2t-2}t\H(t-1)cos(w)-e^{2t-2}\H(t-1)cos(w)