What is (\partial)/(\partial x)((xy-3)e^{-xy}) ?

The answer to (\partial)/(\partial x)((xy-3)e^{-xy}) is -e^{-xy}xy^2+4e^{-xy}y

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(\partial)/(\partial x)((xy-3)e^{-xy})

\frac{\partial}{\partial x} ((x y - 3) e^{- x y})

- e^{- x y} x y^{2} + 4 e^{- x y} y

Solution steps

\frac{\partial}{\partial x} ((x y - 3) e^{- x y})

Treat

y

as a constant

Apply the Product Rule:

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

f = x y - 3, g = e^{- x y}

= \frac{\partial}{\partial x} (x y - 3) e^{- x y} + \frac{\partial}{\partial x} (e^{- x y}) (x y - 3)

\frac{\partial}{\partial x} (x y - 3) = y

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

= y e^{- x y} + (- y e^{- y x}) (x y - 3)

Simplify

y e^{- x y} + (- y e^{- y x}) (x y - 3) : - e^{- x y} x y^{2} + 4 e^{- x y} y

= - e^{- x y} x y^{2} + 4 e^{- x y} y

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 x^{2} - 5 x + 1

\int \frac{4 x ^{2} + 2}{1 + x ^{2}} d x

\int_{1}^{5} \frac{3}{x ^{2}} d x

x^{2} y^{^{''}} - 4 x y^{^{'}} + 6 y = 0

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 3

What is (\partial)/(\partial x)((xy-3)e^{-xy}) ?
The answer to (\partial)/(\partial x)((xy-3)e^{-xy}) is -e^{-xy}xy^2+4e^{-xy}y