What is (\partial)/(\partial x)((3xy+1)e^{-xy}) ?

The answer to (\partial)/(\partial x)((3xy+1)e^{-xy}) is -3e^{-xy}xy^2+2e^{-xy}y

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(\partial)/(\partial x)((3xy+1)e^{-xy})

\frac{\partial}{\partial x} ((3 x y + 1) e^{- x y})

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

Solution steps

\frac{\partial}{\partial x} ((3 x y + 1) e^{- x y})

Treat

y

as a constant

Apply the Product Rule:

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

f = 3 x y + 1, g = e^{- x y}

= \frac{\partial}{\partial x} (3 x y + 1) e^{- x y} + \frac{\partial}{\partial x} (e^{- x y}) (3 x y + 1)

\frac{\partial}{\partial x} (3 x y + 1) = 3 y

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

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

Simplify

3 y e^{- x y} + (- y e^{- y x}) (3 x y + 1) : - 3 e^{- x y} x y^{2} + 2 e^{- x y} y

= - 3 e^{- x y} x y^{2} + 2 e^{- x y} y

4 x^{2} - 15

f (x) = \frac{7}{ln ( x )}

\int \frac{( 4 x + 10 )}{x ^{2} + 2 x + 5} d x

n = 1 \sum \infty \frac{4}{3 ^{n} + 1}

y = 3 x^{2} + 2^{2} x^{2} + 1, (- 1, 0)

What is (\partial)/(\partial x)((3xy+1)e^{-xy}) ?
The answer to (\partial)/(\partial x)((3xy+1)e^{-xy}) is -3e^{-xy}xy^2+2e^{-xy}y