What is (\partial)/(\partial y)(xye^{-1y}) ?

The answer to (\partial)/(\partial y)(xye^{-1y}) is x(-e^{-y}y+e^{-y})

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(\partial)/(\partial y)(xye^{-1y})

\frac{\partial}{\partial y} (x y e^{- 1 y})

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

Solution steps

\frac{\partial}{\partial y} (x y e^{- 1 \cdot y})

Simplify

x y e^{- 1 \cdot y} : e^{- y} x y

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

What is (\partial)/(\partial y)(xye^{-1y}) ?
The answer to (\partial)/(\partial y)(xye^{-1y}) is x(-e^{-y}y+e^{-y})