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

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

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

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

- 3 (e^{- 2 x y} - 2 e^{- 2 x y} x y)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= - 3 \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}

= - 3 (\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}

= - 3 (1 \cdot e^{- 2 x y} + (- 2 y e^{- 2 y x}) x)

Simplify

= - 3 (e^{- 2 x y} - 2 e^{- 2 x y} x y)

\frac{d}{d x} (3 x^{2} - 5 x)

x \to 1 + lim ((x - 2)^{2})

y^{^{''}} + 9 y = 30 sin (3 t)

2 \cdot x \cdot y^{2} + 2 \cdot x + (x^{2} \cdot y + 4 \cdot y) \cdot y^{^{'}} = 0

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

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