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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 0 lim (x^{6} cos (\frac{2}{x}))

x \to 0 lim (2 x + 2)

x \to 1 lim (\frac{e ^{x^{2} - 1}}{x ^{2} - 3 x + 2})

\int (3 cos (θ))^{2} d θ

e^{8 t}

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