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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

= 2 x \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}}

= 2 x \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

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

Simplify

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

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

\int x x^{2} + 2 x + 4 d x

\frac{\partial}{\partial x} (x^{3} sin (5 y))

l a pl a ce t r an s f or m 2 e^{- t} sin (2 t)

l a pl a ce t r an s f or m (1 + t e^{- t})^{3}

\int \frac{2 x}{x - 1} d x

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