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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

4 x^{2} + 5 y \frac{d y}{d x} = 0

x^{^{''}} + x^{^{'}} - 2 x = e^{t}

\int_{0}^{4} π x^{2} d x

- 5 e^{- \frac{x}{3}} + 2

x \to - 4 lim (\frac{x - 4}{x - 8})

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