What is (\partial)/(\partial y)(2ln(x)+ln(y)) ?

The answer to (\partial)/(\partial y)(2ln(x)+ln(y)) is 1/y

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(\partial)/(\partial y)(2ln(x)+ln(y))

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

\frac{1}{y}

Solution steps

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

Simplify

2 ln (x) + ln (y) : ln (x^{2} y)

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

= \frac{1}{x ^{2} y} x^{2}

Simplify

\frac{1}{x ^{2} y} x^{2} : \frac{1}{y}

= \frac{1}{y}

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

\int_{- 1}^{1} e^{x^{2}} d x

\int x^{4} - 2 x^{2} - 4 x d x

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

\frac{d}{d x} ((x^{2} - 2)^{- 3})

What is (\partial)/(\partial y)(2ln(x)+ln(y)) ?
The answer to (\partial)/(\partial y)(2ln(x)+ln(y)) is 1/y