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

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

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

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

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

Solution steps

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

Simplify

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

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

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

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

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

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

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

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