What is the derivative of y=(4x^2-8x+8)e^{-4x} ?

The derivative of y=(4x^2-8x+8)e^{-4x} is -16e^{-4x}x^2+40e^{-4x}x-40e^{-4x}

What is the first derivative of y=(4x^2-8x+8)e^{-4x} ?

The first derivative of y=(4x^2-8x+8)e^{-4x} is -16e^{-4x}x^2+40e^{-4x}x-40e^{-4x}

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derivative of

y = (4 x^{2} - 8 x + 8) e^{- 4 x}

- 16 e^{- 4 x} x^{2} + 40 e^{- 4 x} x - 40 e^{- 4 x}

Solution steps

\frac{d}{d x} ((4 x^{2} - 8 x + 8) e^{- 4 x})

Apply the Product Rule:

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

f = 4 x^{2} - 8 x + 8, g = e^{- 4 x}

= \frac{d}{d x} (4 x^{2} - 8 x + 8) e^{- 4 x} + \frac{d}{d x} (e^{- 4 x}) (4 x^{2} - 8 x + 8)

\frac{d}{d x} (4 x^{2} - 8 x + 8) = 8 x - 8

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

= (8 x - 8) e^{- 4 x} + (- 4 e^{- 4 x}) (4 x^{2} - 8 x + 8)

Simplify

(8 x - 8) e^{- 4 x} + (- 4 e^{- 4 x}) (4 x^{2} - 8 x + 8) : - 16 e^{- 4 x} x^{2} + 40 e^{- 4 x} x - 40 e^{- 4 x}

= - 16 e^{- 4 x} x^{2} + 40 e^{- 4 x} x - 40 e^{- 4 x}

\frac{\partial}{\partial x} (ln (x + 9 y + 8 z))

\frac{\partial}{\partial x} (x y^{2} z^{4})

\int_{2}^{1} - 4 x^{- 3} d x

\int \frac{x ^{7}}{( 2 - x ^{8} ) ^{2}} d x

10000 y^{^{''}} - 81 y = 0, y (0) = 1, y^{^{'}} (0) = 3

What is the derivative of y=(4x^2-8x+8)e^{-4x} ?
The derivative of y=(4x^2-8x+8)e^{-4x} is -16e^{-4x}x^2+40e^{-4x}x-40e^{-4x}
What is the first derivative of y=(4x^2-8x+8)e^{-4x} ?
The first derivative of y=(4x^2-8x+8)e^{-4x} is -16e^{-4x}x^2+40e^{-4x}x-40e^{-4x}