What is (\partial)/(\partial y)(xy+ln(x/y)) ?

The answer to (\partial)/(\partial y)(xy+ln(x/y)) is x-1/y

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(\partial)/(\partial y)(xy+ln(x/y))

\frac{\partial}{\partial y} (x y + ln (\frac{x}{y}))

x - \frac{1}{y}

Solution steps

\frac{\partial}{\partial y} (x y + ln (\frac{x}{y}))

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x y) + \frac{\partial}{\partial y} (ln (\frac{x}{y}))

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

\frac{\partial}{\partial y} (ln (\frac{x}{y})) = - \frac{1}{y}

= x - \frac{1}{y}

\int x^{4} d x

\int - x^{- 2} d x

\int \frac{1}{8 + 5 x} d x

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

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

What is (\partial)/(\partial y)(xy+ln(x/y)) ?
The answer to (\partial)/(\partial y)(xy+ln(x/y)) is x-1/y