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

The answer to (\partial)/(\partial y)(y-y/(x^2+y^2)) is 1-(x^2-y^2)/((x^2+y^2)^2)

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

\frac{d N}{d x} = ((0.0922) N) (1 - \frac{N}{60})

\frac{d}{d x} ((e^{x} - e^{- 2 x})^{\frac{1}{5}})

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

\int 4 cos (x) d x

f (x) = x^{\frac{3}{2}} + \frac{1}{x ^{\frac{5}{2}}}

What is (\partial)/(\partial y)(y-y/(x^2+y^2)) ?
The answer to (\partial)/(\partial y)(y-y/(x^2+y^2)) is 1-(x^2-y^2)/((x^2+y^2)^2)