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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{\partial}{\partial y} (y) = 1

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

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

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

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

\int e^{2 t} d t

\int \frac{9 - x ^{2}}{2 x ^{2}} d x

(x^{2} - 2 x + 2) e^{\frac{5 x}{2}}

x^{2} + y^{2} = 26

\frac{d}{d x} (- x^{3} + 5 x^{2} - 8 x)

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