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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

x (\frac{d y}{d x}) = x^{2} - 4

f (x) = (5 x^{2} - 7 x) e^{x}

f (x) = ln^{3} (x)

\frac{d}{d x} (2 - \frac{2}{3} x)

\int \frac{1}{5} x^{5} d x

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