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}})

Treat

x

as a constant

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}}

\int (4 x^{3} - 3 x + 5) d x

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

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

e^{\frac{t}{3}}

t sin (0.32 t)

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)