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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = e^{- x - y}, g = sin (y)

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

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

\frac{\partial}{\partial y} (sin (y)) = cos (y)

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

Simplify

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

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

f (x) = (5 x^{3} + 2)^{4}

\frac{d}{d x} (tan (arcsin (x)))

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

\int \frac{( 5 x - 12 )}{x ^{3} - 6 x ^{2} + 8 x} d x

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