What is the d/(dy)(1/y e^{xy}-ln(y^2)) ?

The d/(dy)(1/y e^{xy}-ln(y^2)) is (xe^{xy}y-e^{xy})/(y^2)-2/y

What is the first d/(dy)(1/y e^{xy}-ln(y^2)) ?

The first d/(dy)(1/y e^{xy}-ln(y^2)) is (xe^{xy}y-e^{xy})/(y^2)-2/y

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\frac{d}{d y} (\frac{1}{y} e^{x y} - ln (y^{2}))

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

Solution steps

\frac{d}{d y} (\frac{1}{y} e^{x y} - ln (y^{2}))

Simplify

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

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d y} (ln (y^{2})) = \frac{2}{y}

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

x \to - \infty lim (2 x - 3)

\int 3 x^{2} + y^{2} d x

x \to 6 lim (\frac{1}{x})

\frac{d x}{d t} = x^{\frac{1}{2}}

f (5)

What is the d/(dy)(1/y e^{xy}-ln(y^2)) ?
The d/(dy)(1/y e^{xy}-ln(y^2)) is (xe^{xy}y-e^{xy})/(y^2)-2/y
What is the first d/(dy)(1/y e^{xy}-ln(y^2)) ?
The first d/(dy)(1/y e^{xy}-ln(y^2)) is (xe^{xy}y-e^{xy})/(y^2)-2/y