What is expand 1/8 e^{4x}(-4e^{-4t}t-e^{-4t}) ?

The solution to expand 1/8 e^{4x}(-4e^{-4t}t-e^{-4t}) is -(e^{4x-4t}t)/2-1/8 e^{4x-4t}

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expand 1/8 e^{4x}(-4e^{-4t}t-e^{-4t})

expand

\frac{1}{8} e^{4 x} (- 4 e^{- 4 t} t - e^{- 4 t})

- \frac{e ^{4 x - 4 t} t}{2} - \frac{1}{8} e^{4 x - 4 t}

Solution steps

\frac{1}{8} e^{4 x} (- 4 e^{- 4 t} t - e^{- 4 t})

Apply the distributive law:

a (b - c) = ab - a c

a = \frac{1}{8} e^{4 x}, b = - 4 e^{- 4 t} t, c = e^{- 4 t}

= \frac{1}{8} e^{4 x} (- 4 e^{- 4 t} t) - \frac{1}{8} e^{4 x} e^{- 4 t}

Apply minus-plus rules

+ (- a) = - a

= - 4 \cdot \frac{1}{8} e^{4 x} e^{- 4 t} t - \frac{1}{8} e^{4 x} e^{- 4 t}

Simplify

- 4 \cdot \frac{1}{8} e^{4 x} e^{- 4 t} t - \frac{1}{8} e^{4 x} e^{- 4 t} : - \frac{e ^{4 x - 4 t} t}{2} - \frac{1}{8} e^{4 x - 4 t}

= - \frac{e ^{4 x - 4 t} t}{2} - \frac{1}{8} e^{4 x - 4 t}

2 cos (5) (2 tan (5) x + 25)

y^{2} d x + (x^{2} + x y) d y

sin (θ) - cos (θ) (sin (θ) - cos (θ))

(i + 2)^{4}

csc (a) (tan (a) - sin (a))

What is expand 1/8 e^{4x}(-4e^{-4t}t-e^{-4t}) ?
The solution to expand 1/8 e^{4x}(-4e^{-4t}t-e^{-4t}) is -(e^{4x-4t}t)/2-1/8 e^{4x-4t}