What is expand te^{4t}f(t-3)\H(t-3) ?

The solution to expand te^{4t}f(t-3)\H(t-3) is e^{4t}t^2f\H(t-3)-3e^{4t}tf\H(t-3)

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expand te^{4t}f(t-3)\H(t-3)

expand

t e^{4 t} f (t - 3) H (t - 3)

e^{4 t} t^{2} f H (t - 3) - 3 e^{4 t} t f H (t - 3)

Solution steps

t e^{4 t} f (t - 3) H (t - 3)

= e^{4 t} t f H (t - 3) (t - 3)

Apply the distributive law:

a (b - c) = ab - a c

a = t e^{4 t} f H (t - 3), b = t, c = 3

= t e^{4 t} f H (t - 3) t - t e^{4 t} f H (t - 3) \cdot 3

= e^{4 t} tt f H (t - 3) - 3 e^{4 t} t f H (t - 3)

e^{4 t} tt f H (t - 3) = e^{4 t} t^{2} f H (t - 3)

= e^{4 t} t^{2} f H (t - 3) - 3 e^{4 t} t f H (t - 3)

e^{2 (s i n (t) + 1)}

\frac{3}{5 ( 2 x - 4 ) ^{2}}

\frac{2 s ^{2} + 19 s + 40}{( s + 3 ) ( s ^{2} + 6 s + 9 )}

\frac{20}{( x + 1 ) ^{3}}

l im x \Rightarrow 3 (5 x - lo g_{10} (x))

What is expand te^{4t}f(t-3)\H(t-3) ?
The solution to expand te^{4t}f(t-3)\H(t-3) is e^{4t}t^2f\H(t-3)-3e^{4t}tf\H(t-3)