What is the derivative of 4x^2e^{-0.25x} ?

The derivative of 4x^2e^{-0.25x} is 4(2e^{-0.25x}x-0.25e^{-0.25x}x^2)

What is the first derivative of 4x^2e^{-0.25x} ?

The first derivative of 4x^2e^{-0.25x} is 4(2e^{-0.25x}x-0.25e^{-0.25x}x^2)

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\frac{d}{d x} (4 x^{2} e^{- 0.25 x})

4 (2 e^{- 0.25 x} x - 0.25 e^{- 0.25 x} x^{2})

Solution steps

\frac{d}{d x} (4 x^{2} e^{- 0.25 x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d x} (x^{2} e^{- 0.25 x})

Apply the Product Rule:

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

f = x^{2}, g = e^{- 0.25 x}

= 4 (\frac{d}{d x} (x^{2}) e^{- 0.25 x} + \frac{d}{d x} (e^{- 0.25 x}) x^{2})

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

\frac{d}{d x} (e^{- 0.25 x}) = - 0.25 e^{- 0.25 x}

= 4 (2 x e^{- 0.25 x} + (- 0.25 e^{- 0.25 x}) x^{2})

Simplify

= 4 (2 e^{- 0.25 x} x - 0.25 e^{- 0.25 x} x^{2})

(x^{2} + e^{x}) (x)

\int \frac{x ^{2} + 60 x + 60}{x ^{3} - 4 x} d x

n = 1 \sum \infty (- 0.3)^{n - 1}

x \to 3 + lim (\frac{5 x}{x ^{2} - 9})

\frac{d}{d x} (ln (cosh (2 x)))

What is the derivative of 4x^2e^{-0.25x} ?
The derivative of 4x^2e^{-0.25x} is 4(2e^{-0.25x}x-0.25e^{-0.25x}x^2)
What is the first derivative of 4x^2e^{-0.25x} ?
The first derivative of 4x^2e^{-0.25x} is 4(2e^{-0.25x}x-0.25e^{-0.25x}x^2)