What is (\partial)/(\partial x)((2xy-3)e^{-xy}) ?

The answer to (\partial)/(\partial x)((2xy-3)e^{-xy}) is -2e^{-xy}xy^2+5e^{-xy}y

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(\partial)/(\partial x)((2xy-3)e^{-xy})

\frac{\partial}{\partial x} ((2 x y - 3) e^{- x y})

- 2 e^{- x y} x y^{2} + 5 e^{- x y} y

Solution steps

\frac{\partial}{\partial x} ((2 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 = 2 x y - 3, g = e^{- x y}

= \frac{\partial}{\partial x} (2 x y - 3) e^{- x y} + \frac{\partial}{\partial x} (e^{- x y}) (2 x y - 3)

\frac{\partial}{\partial x} (2 x y - 3) = 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 - 3)

Simplify

2 y e^{- x y} + (- y e^{- y x}) (2 x y - 3) : - 2 e^{- x y} x y^{2} + 5 e^{- x y} y

= - 2 e^{- x y} x y^{2} + 5 e^{- x y} y

\int x^{5} e^{(x^{6} - 4)} d x

\int \frac{x ^{4}}{2} + cos (x) x d x

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

f (x) = 5 x - 4 x, (1, 1)

sin (x) y^{^{'}} + cos (x) y = 0

What is (\partial)/(\partial x)((2xy-3)e^{-xy}) ?
The answer to (\partial)/(\partial x)((2xy-3)e^{-xy}) is -2e^{-xy}xy^2+5e^{-xy}y