What is (\partial)/(\partial x)(xe^{x-y}) ?

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

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(\partial)/(\partial x)(xe^{x-y})

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

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

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

Simplify

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

\frac{d}{d x} (\frac{4 x ^{3} - 7 x + 8}{x})

\frac{d y}{d x} = \frac{1 + 3 x ^{2}}{( 1 + y ) ^{2}}, y (1) = 2

\int e^{- s \cdot t} d t

t^{2} y^{^{''}} + 2 t y^{^{'}} - 2 y = 0

\int \frac{4 x + 3}{2 x + 1} d x

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