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

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

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

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

\int \frac{1}{x 3 x} d x

\frac{d}{d x} (2 ln (x^{2} + y^{2}))

t y^{^{'}} = (1 - y^{2})^{\frac{1}{2}}

\int \frac{e ^{2 x}}{x ^{2}} d x

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