What is (\partial)/(\partial x)(8xye^{-x^2-y^2}) ?

The answer to (\partial)/(\partial x)(8xye^{-x^2-y^2}) is 8y(e^{-x^2-y^2}-2e^{-x^2-y^2}x^2)

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(\partial)/(\partial x)(8xye^{-x^2-y^2})

\frac{\partial}{\partial x} (8 x y e^{- x^{2} - y^{2}})

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

Solution steps

\frac{\partial}{\partial x} (8 x y e^{- x^{2} - y^{2}})

Treat

y

as a constant

Take the constant out:

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

= 8 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}}

= 8 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

= 8 y (1 \cdot e^{- x^{2} - y^{2}} + (- 2 e^{- x^{2} - y^{2}} x) x)

Simplify

8 y (1 \cdot e^{- x^{2} - y^{2}} + (- 2 e^{- x^{2} - y^{2}} x) x) : 8 y (e^{- x^{2} - y^{2}} - 2 e^{- x^{2} - y^{2}} x^{2})

= 8 y (e^{- x^{2} - y^{2}} - 2 e^{- x^{2} - y^{2}} x^{2})

What is (\partial)/(\partial x)(8xye^{-x^2-y^2}) ?
The answer to (\partial)/(\partial x)(8xye^{-x^2-y^2}) is 8y(e^{-x^2-y^2}-2e^{-x^2-y^2}x^2)