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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (6 sec (x) tan (x))

\int_{0}^{\infty} e^{\frac{x}{3}} d x

x^{2} \frac{d y}{d x} + 4 x y - y^{3} = 0

\int_{0}^{1} 2 π (x) (8 - 7 x - x^{2}) d x

\int_{0}^{1} \frac{5}{x ^{5}} d x

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