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

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

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

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

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

Solution steps

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

Take the constant out:

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

= - 2 \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}}

= - 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

= - 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})

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0

sec^{2} (x)

2 y^{^{''}} + 17 y^{^{'}} - 9 y = 0

\frac{d}{d y} (x y + y z + x z)

y^{^{'}} = 3 e^{x - y}

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