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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\int sin^{2} (y) d y

y = 7 x + 4, (3, 5)

\int_{0}^{1} [x - x^{2}] d x

9 x + 4

x \to - 6 - lim (\frac{- 9 x}{x + 6})

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