What is (\partial)/(\partial x)((xy+x+y)e^x) ?

The answer to (\partial)/(\partial x)((xy+x+y)e^x) is 2e^xy+e^xxy+e^x+e^xx

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(\partial)/(\partial x)((xy+x+y)e^x)

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

g (x) = \frac{( x ^{2} - 1 )}{x ^{2} + x + 1}

x \to 0 lim (2 x - p (x))

\frac{d}{d x} (\frac{π x}{12})

6 \frac{d y}{d x} + 3 e^{x + y} = 0

g (x) = x, (1, 1)

What is (\partial)/(\partial x)((xy+x+y)e^x) ?
The answer to (\partial)/(\partial x)((xy+x+y)e^x) is 2e^xy+e^xxy+e^x+e^xx