What is (\partial)/(\partial y)(1+ln(xy)) ?

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

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(\partial)/(\partial y)(1+ln(xy))

\frac{\partial}{\partial y} (1 + ln (x y))

\frac{1}{y}

Solution steps

\frac{\partial}{\partial y} (1 + ln (x y))

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (1) = 0

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

= 0 + \frac{1}{y}

Simplify

= \frac{1}{y}

\int_{- 1}^{1} \frac{3}{x ^{2}} d x

\frac{\partial}{\partial x} (L + (L^{2} + x^{2})^{\frac{1}{2}})

\frac{d}{d y} (29 - x^{2} - 4 y^{2})

\int (\frac{x}{x - 1}) d x

x^{3} - 25 x + 5, at 5

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