What is the derivative of 2xe^{-x^2} ?

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

What is the first derivative of 2xe^{-x^2} ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

(y^{16}) d y = \frac{( 1 + x )}{x} d x, y (1) = 5

\frac{( s ^{2} + s )}{2 s ^{2} + 2 s + 1}

f (x) = \frac{1}{( 3 x + 8 ) ^{2}}

\frac{d}{d x} (\frac{- x ^{2} + 2 x + 1}{( x ^{2} + 1 ) ^{2}})

\int cos (2 πn x) d x

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