What is expand-2(e^xsin(x)+cos(x)e^x) ?

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

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expand-2(e^xsin(x)+cos(x)e^x)

expand

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

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

Solution steps

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

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

Apply the distributive law:

a (b + c) = ab + a c

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

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

Apply minus-plus rules

+ (- a) = - a

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

cos (x + 1) (1 + \frac{d y}{d x})

16 e^{4 x} (1 + e^{4 x})

\frac{( x ^{3} - 2 ) ^{9}}{x ^{18}}

(d - e)^{3}

(1, 0, - 3) x (5, 2, 6)

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