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

The answer to (\partial)/(\partial x)(xye^{-xy}) is y(e^{-xy}-e^{-xy}xy)

(\partial)/(\partial x)(xye^{-xy})

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

\frac{d}{d x} (ln (3 x + 2 y))

y^{^{''}} + 7 y^{^{'}} + 23 y = 0

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

\int \frac{2 z - 3}{7 z - 10} d z

\frac{\partial}{\partial x} (2 sin (x) cos (y))

What is (\partial)/(\partial x)(xye^{-xy}) ?
The answer to (\partial)/(\partial x)(xye^{-xy}) is y(e^{-xy}-e^{-xy}xy)