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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\int_{0}^{4} x 16 - x^{2} d x

\int \frac{1 - t}{t} d t

x \to 1 lim (3 x^{3} - 5 x^{2} + x + 5)

\frac{d}{d x} (- \frac{1}{5} cos (5 x))

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = - 3

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