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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

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

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

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

\frac{\partial}{\partial x} (e^{- \frac{y}{x}}) = \frac{y e ^{- \frac{y}{x}}}{x ^{2}}

= 1 \cdot e^{- \frac{y}{x}} + \frac{y e ^{- \frac{y}{x}}}{x ^{2}} x

Simplify

1 \cdot e^{- \frac{y}{x}} + \frac{y e ^{- \frac{y}{x}}}{x ^{2}} x : e^{- \frac{y}{x}} + \frac{e ^{- \frac{y}{x}} y}{x}

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

f (t) = \frac{3}{5} e^{- t} - \frac{22}{5} e^{4 t}

\int (\frac{3}{x ^{2}} - 1) d x

\frac{d y}{d x} = 2 - y

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 2

\int \frac{1}{9 x ^{2} + 5} d x

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