What is expand (e^xcos(x)-sin(x)e^x)(-2e^xsin(x)) ?

The solution to expand (e^xcos(x)-sin(x)e^x)(-2e^xsin(x)) is -2e^{2x}cos(x)sin(x)+2e^{2x}sin^2(x)

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expand (e^xcos(x)-sin(x)e^x)(-2e^xsin(x))

expand

(e^{x} cos (x) - sin (x) e^{x}) (- 2 e^{x} sin (x))

- 2 e^{2 x} cos (x) sin (x) + 2 e^{2 x} sin^{2} (x)

Solution steps

(e^{x} cos (x) - sin (x) e^{x}) (- 2 e^{x} sin (x))

= (e^{x} cos (x) - e^{x} sin (x)) (- 2 e^{x} sin (x))

Apply the distributive law:

a (b - c) = ab - a c

a = (- 2 e^{x} sin (x)), b = e^{x} cos (x), c = sin (x) e^{x}

= (- 2 e^{x} sin (x)) e^{x} cos (x) - (- 2 e^{x} sin (x)) sin (x) e^{x}

= e^{x} cos (x) (- 2 e^{x} sin (x)) - e^{x} sin (x) (- 2 e^{x} sin (x))

Simplify

e^{x} cos (x) (- 2 e^{x} sin (x)) - e^{x} sin (x) (- 2 e^{x} sin (x)) : - 2 e^{2 x} cos (x) sin (x) + 2 e^{2 x} sin^{2} (x)

= - 2 e^{2 x} cos (x) sin (x) + 2 e^{2 x} sin^{2} (x)

\frac{( x + 1 )}{2 x ( 2 x + 1 )}

(x - y)^{2} ((x - y) - Δ)

2 u y^{2} (y d u + u d y) + (y^{2} - 3 y^{2} v^{2}) d u

\frac{4 ( 3 ( 3 a + b ) - ( 5 a - b ) )}{12}

, (x - 1) (x - 2) (x - 4)

What is expand (e^xcos(x)-sin(x)e^x)(-2e^xsin(x)) ?
The solution to expand (e^xcos(x)-sin(x)e^x)(-2e^xsin(x)) is -2e^{2x}cos(x)sin(x)+2e^{2x}sin^2(x)