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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

\frac{\partial}{\partial x} (x) = 1

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

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

Simplify

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

\frac{d ^{2}}{d x ^{2}} (θ cos (θ))

x \to 0 lim (1 - \frac{x ^{2}}{4})

x \to 4 lim (16 - x^{2})

g (x) = \frac{3 x}{x - 2}, (4, 6)

y = 8 x

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