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

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

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

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

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

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

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

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

Simplify

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

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

\int - e^{x} x d x

\int \frac{36}{1 + 36 x ^{2}} d x

y^{^{'}} = 1 - 2 y

f (x) = \frac{x}{1 - ln ( x - 4 )}

f (x) = (1 + 3 x) (5 - 4 x), at x = 1

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