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

The answer to (\partial)/(\partial y)((2x)/(x^2+y^2)) is -(4xy)/((x^2+y^2)^2)

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

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

- \frac{4 x y}{( x ^{2} + y ^{2} ) ^{2}}

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

2 x (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 y) : - \frac{4 x y}{( x ^{2} + y ^{2} ) ^{2}}

= - \frac{4 x y}{( x ^{2} + y ^{2} ) ^{2}}

\int \frac{x ^{2}}{( 16 + x ^{2} ) ^{2}} d x

\int (8 x - 7) \cdot ln (x) d x

x 8 - 2 x^{2}, 0, 2

\frac{\partial}{\partial x} (ln (\frac{x}{y}))

What is (\partial)/(\partial y)((2x)/(x^2+y^2)) ?
The answer to (\partial)/(\partial y)((2x)/(x^2+y^2)) is -(4xy)/((x^2+y^2)^2)