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)

d er i v a t i v e e^{\frac{3}{x}}

\int cos^{3} (\frac{x}{13}) d x

d er i v a t i v e y = (9 x^{4} + 5 x)^{3}

\int e^{8 x + 7} d x

\frac{d}{d x} (\frac{4 x ^{3} + 9}{x})

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