What is the derivative of-8e^{-4x^2}x ?

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

What is the first derivative of-8e^{-4x^2}x ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

\frac{d x}{d x} = 1

= - 8 ((- 8 e^{- 4 x^{2}} x) x + 1 \cdot e^{- 4 x^{2}})

Simplify

- 8 ((- 8 e^{- 4 x^{2}} x) x + 1 \cdot e^{- 4 x^{2}}) : - 8 (- 8 e^{- 4 x^{2}} x^{2} + e^{- 4 x^{2}})

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

\int \frac{1}{1 + e ^{- y}} d y

f (x) = (x + x^{- 1})^{3}

\int tan^{3} (5 x) d x

y = (ln ∣ x + 8 ∣)^{7}

y^{^{'}} = \frac{( 3 y ^{2} - 9 )}{2 x y}

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