What is (\partial)/(\partial y)(-arctan(x/y)) ?

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

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(\partial)/(\partial y)(-arctan(x/y))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - \frac{\partial}{\partial y} (arctan (\frac{x}{y}))

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int (e^{x} + 1) d x

\frac{d}{d x} (x^{4} + 12 x^{2} + 42)

\frac{d}{d x} (\frac{6 x}{2 + x ^{2}})

x y^{^{'}} = 2 y + x^{3}

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

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