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

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

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

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

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

Solution steps

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

Simplify

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

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

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

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

\frac{d}{d x} (\frac{3 x - 1}{x ^{2} + 2})

\frac{d}{d x} (sin (8 cos (x)))

\int \frac{tan ^{3} ( θ )}{cos ^{4} ( θ )} d θ

x y^{^{''}} + y^{^{'}} = 0

\frac{d}{d x} (π^{x})

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