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

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

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

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\int_{0}^{1} \frac{7}{x ( x + 1 )} d x

\int_{3}^{5} \frac{1}{x - 4} d x

y = (x^{2} - 4 x)^{5}

\int_{1}^{\infty} e^{- x} d x

(x^{x^{x^{x}}})^{^{'}}

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