What is the u=(x^2+4x+4)e^{-2x} ?

The u=(x^2+4x+4)e^{-2x} is -2e^{-2x}x^2-6e^{-2x}x-4e^{-2x}

What is the first u=(x^2+4x+4)e^{-2x} ?

The first u=(x^2+4x+4)e^{-2x} is -2e^{-2x}x^2-6e^{-2x}x-4e^{-2x}

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\frac{d}{d x} ((x^{2} + 4 x + 4) e^{- 2 x})

- 2 e^{- 2 x} x^{2} - 6 e^{- 2 x} x - 4 e^{- 2 x}

Solution steps

\frac{d}{d x} ((x^{2} + 4 x + 4) e^{- 2 x})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} + 4 x + 4, g = e^{- 2 x}

= \frac{d}{d x} (x^{2} + 4 x + 4) e^{- 2 x} + \frac{d}{d x} (e^{- 2 x}) (x^{2} + 4 x + 4)

\frac{d}{d x} (x^{2} + 4 x + 4) = 2 x + 4

\frac{d}{d x} (e^{- 2 x}) = - 2 e^{- 2 x}

= (2 x + 4) e^{- 2 x} + (- 2 e^{- 2 x}) (x^{2} + 4 x + 4)

Simplify

(2 x + 4) e^{- 2 x} + (- 2 e^{- 2 x}) (x^{2} + 4 x + 4) : - 2 e^{- 2 x} x^{2} - 6 e^{- 2 x} x - 4 e^{- 2 x}

= - 2 e^{- 2 x} x^{2} - 6 e^{- 2 x} x - 4 e^{- 2 x}

y = 0.6 sin (\frac{π}{6} x + 1) + 0.8

y = (x^{3} - x)^{3}

h (x) = - 4 2 x - 8

f (x) = (ln (4 x))^{2}

f (x) = 100 - x

What is the u=(x^2+4x+4)e^{-2x} ?
The u=(x^2+4x+4)e^{-2x} is -2e^{-2x}x^2-6e^{-2x}x-4e^{-2x}
What is the first u=(x^2+4x+4)e^{-2x} ?
The first u=(x^2+4x+4)e^{-2x} is -2e^{-2x}x^2-6e^{-2x}x-4e^{-2x}