What is (\partial)/(\partial y)(1/y e^{xy}) ?

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

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(\partial)/(\partial y)(1/y e^{xy})

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

- \frac{e ^{y x}}{y ^{2}} + \frac{e ^{y x} x}{y}

Solution steps

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

Treat

x

as a constant

Apply the Product Rule:

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

f = \frac{1}{y}, g = e^{x y}

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

\frac{\partial}{\partial y} (\frac{1}{y}) = - \frac{1}{y ^{2}}

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

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

Simplify

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

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

\int_{1}^{13} \frac{x - 1}{72} d x

(- 1, 1), (1, 0)

\frac{d}{d x} (\frac{2}{cos ( 7 x )})

f (x) = x^{4}

\frac{d}{d t} (- 5 sin (t))

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