What is the derivative of y=-2e^{-x^2}x ?

The derivative of y=-2e^{-x^2}x is -2(-2e^{-x^2}x^2+e^{-x^2})

What is the first derivative of y=-2e^{-x^2}x ?

The first derivative of y=-2e^{-x^2}x is -2(-2e^{-x^2}x^2+e^{-x^2})

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\int \frac{3 x}{1 + x ^{3}} d x

y^{^{'}} + 2 x y = y + 4 x - 2

\frac{d}{d x} (\frac{1}{1 + e ^{x}})

\frac{\partial}{\partial y} (3 x^{2} + y^{2})

3 x y^{^{'}} + 4 y = \frac{x}{y}

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