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

The answer to (\partial)/(\partial x)(xe^{-2x^2-2y^2}) is e^{-2x^2-2y^2}-4e^{-2x^2-2y^2}x^2

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

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

e^{- 2 x^{2} - 2 y^{2}} - 4 e^{- 2 x^{2} - 2 y^{2}} x^{2}

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

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

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

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

\frac{\partial}{\partial x} (e^{- 2 x^{2} - 2 y^{2}}) = - 4 e^{- 2 x^{2} - 2 y^{2}} x

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

Simplify

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

= e^{- 2 x^{2} - 2 y^{2}} - 4 e^{- 2 x^{2} - 2 y^{2}} x^{2}

x^{2} y^{^{''}} - 9 x y^{^{'}} + 25 y = 0

d er i v a t i v e f (x) = 36 x^{- 2} - 2 x^{- 4} + 2

\frac{d}{d x} ((x + b)^{2})

d er i v a t i v e a^{7} + cos^{7} (x)

\frac{d}{d x} (\frac{- 7 x}{2 x - 1})

What is (\partial)/(\partial x)(xe^{-2x^2-2y^2}) ?
The answer to (\partial)/(\partial x)(xe^{-2x^2-2y^2}) is e^{-2x^2-2y^2}-4e^{-2x^2-2y^2}x^2