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

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

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

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

\frac{1}{x + y}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{1}{x + y}

\frac{d}{d x} ((f (x) (6 x))^{2} - 4)

\frac{\partial}{\partial x} (x^{3} - 5 x^{2} y + 7 x y^{3})

\int \frac{1}{2 - e ^{- x}} d x

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

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