What is (\partial}{\partial y}(ln(\frac{x^2)/y)) ?

The answer to (\partial}{\partial y}(ln(\frac{x^2)/y)) is -1/y

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(\partial}{\partial y}(ln(\frac{x^2)/y))

\frac{\partial}{\partial y} (ln (\frac{x ^{2}}{y}))

- \frac{1}{y}

Solution steps

\frac{\partial}{\partial y} (ln (\frac{x ^{2}}{y}))

Treat

x

as a constant

Apply the chain rule:

\frac{1}{\frac{x ^{2}}{y}} \frac{\partial}{\partial y} (\frac{x ^{2}}{y})

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

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

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

Simplify

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

= - \frac{1}{y}

\frac{d}{d x} (\frac{4}{9} x^{3})

x sin (x^{2})

\int x - 2 x d x

4 x y

\int (8 x + 13) e^{x} d x

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