What is (\partial)/(\partial x)(xe^{-x^2}) ?

The answer to (\partial)/(\partial x)(xe^{-x^2}) is e^{-x^2}-2e^{-x^2}x^2

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(\partial)/(\partial x)(xe^{-x^2})

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

e^{- x^{2}} - 2 e^{- x^{2}} x^{2}

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{\partial}{\partial x} (x) = 1

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

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

Simplify

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

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

\frac{1}{s + 1 ( - 2 s + 10 )}

n = 1 \sum \infty \frac{( - 5 ) ^{n}}{3 ^{2 n + 1}}

\frac{d}{d x} (32 cos (x))

n = 1 \sum \infty \frac{x ^{n}}{1 6 ^{n}}

f (x) = - \frac{4 x}{x ^{2} + 1}, at x = 1

What is (\partial)/(\partial x)(xe^{-x^2}) ?
The answer to (\partial)/(\partial x)(xe^{-x^2}) is e^{-x^2}-2e^{-x^2}x^2