What is the d/(dy)(1/y e^{xy}) ?

The d/(dy)(1/y e^{xy}) is -(e^{yx}}{y^2}+\frac{e^{yx}x)/y

What is the first d/(dy)(1/y e^{xy}) ?

The first d/(dy)(1/y e^{xy}) is -(e^{yx}}{y^2}+\frac{e^{yx}x)/y

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

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

Solution steps

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

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

\frac{d}{d 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}

y^{2} = (x y y^{^{'}} + 1) ln (x)

y^{^{'}} = 1 - y

t^{2} y^{^{'}} + 6 t y - y^{3} = 0

y^{^{''}} - 2 y^{^{'}} + 4 y = 0

e^{x^{2} - 4 x}

What is the d/(dy)(1/y e^{xy}) ?
The d/(dy)(1/y e^{xy}) is -(e^{yx}}{y^2}+\frac{e^{yx}x)/y
What is the first d/(dy)(1/y e^{xy}) ?
The first d/(dy)(1/y e^{xy}) is -(e^{yx}}{y^2}+\frac{e^{yx}x)/y