What is (\partial)/(\partial x)(xe^{-x}(y^2-4y)) ?

The answer to (\partial)/(\partial x)(xe^{-x}(y^2-4y)) is (y^2-4y)(e^{-x}-e^{-x}x)

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(\partial)/(\partial x)(xe^{-x}(y^2-4y))

\frac{\partial}{\partial x} (x e^{- x} (y^{2} - 4 y))

(y^{2} - 4 y) (e^{- x} - e^{- x} x)

Solution steps

\frac{\partial}{\partial x} (x e^{- x} (y^{2} - 4 y))

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

= (y^{2} - 4 y) (\frac{\partial}{\partial x} (x) e^{- x} + \frac{\partial}{\partial x} (e^{- x}) x)

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

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

= (y^{2} - 4 y) (1 \cdot e^{- x} + (- e^{- x}) x)

Simplify

= (y^{2} - 4 y) (e^{- x} - e^{- x} x)

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + a ^{2}})

\frac{x ^{2}}{3 ( 1 - x )}

\frac{7}{s ^{2} + 5 s + 10}

(20 + x^{2}) y^{^{'}} - 2 x y = 0

What is (\partial)/(\partial x)(xe^{-x}(y^2-4y)) ?
The answer to (\partial)/(\partial x)(xe^{-x}(y^2-4y)) is (y^2-4y)(e^{-x}-e^{-x}x)