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

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

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

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

- \frac{2 y}{x ^{2} + y ^{2}}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int t^{3} 1 + t^{2} d t

\int x 4^{x} d x

y = (cos (x))^{s i n (x)}

\int \frac{x}{2 x + 1} d x

x \to 2 lim ((3 - x) (x + 1))

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