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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

sin (5 t)

y^{^{''}} - y^{^{'}} - 2 y = 8 e - t

f (x) = x^{\frac{1}{x}}

\frac{\partial}{\partial y} (y + y cos (x y))

\frac{d r}{d x} = 4 π r^{2}

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