What is (\partial)/(\partial x)((2xy-1)e^{-xy}) ?

The answer to (\partial)/(\partial x)((2xy-1)e^{-xy}) is -2e^{-xy}xy^2+3e^{-xy}y

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(\partial)/(\partial x)((2xy-1)e^{-xy})

\frac{\partial}{\partial x} ((2 x y - 1) e^{- x y})

- 2 e^{- x y} x y^{2} + 3 e^{- x y} y

Solution steps

\frac{\partial}{\partial x} ((2 x y - 1) e^{- x y})

Treat

y

as a constant

Apply the Product Rule:

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

f = 2 x y - 1, g = e^{- x y}

= \frac{\partial}{\partial x} (2 x y - 1) e^{- x y} + \frac{\partial}{\partial x} (e^{- x y}) (2 x y - 1)

\frac{\partial}{\partial x} (2 x y - 1) = 2 y

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

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

Simplify

2 y e^{- x y} + (- y e^{- y x}) (2 x y - 1) : - 2 e^{- x y} x y^{2} + 3 e^{- x y} y

= - 2 e^{- x y} x y^{2} + 3 e^{- x y} y

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

(9 + y^{2}) d x + 4 x y d y = 0

y = 8 x + 12 - 15 e^{x}, (0, - 3)

\int \frac{3 + x}{1 + x ^{2}} d x

x^{2} + 10, 9 x, 0

What is (\partial)/(\partial x)((2xy-1)e^{-xy}) ?
The answer to (\partial)/(\partial x)((2xy-1)e^{-xy}) is -2e^{-xy}xy^2+3e^{-xy}y