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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = x, g = e^{- 3 x y}

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

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

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

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

Simplify

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

\frac{\partial}{\partial x} (y ln (x^{2} + y^{2} + 3) - 6)

\frac{\partial}{\partial y} (3 x cos (y))

\int sin^{3} (\frac{2}{3} x) cos^{4} (\frac{2}{3} x) d x

\frac{\partial}{\partial x} (2 ln (\frac{3 x y}{5 z}))

y = \frac{2 x}{sin ( x )}

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