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

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

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

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

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

Solution steps

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

Simplify

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

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (x y^{2} - x^{2} y + x^{2} + y^{2})

x \to - 6 lim (\frac{x + 1 + 5}{x + 6})

x \to 9 lim (\frac{2 x - 1}{x + 3})

\frac{\partial}{\partial z} (x \cdot cos (y) \cdot sin (z))

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