What is the d/(dy)((2xy)/((x^2+y^2)^2)) ?

The d/(dy)((2xy)/((x^2+y^2)^2)) is (2x(x^2-3y^2))/((x^2+y^2)^3)

What is the first d/(dy)((2xy)/((x^2+y^2)^2)) ?

The first d/(dy)((2xy)/((x^2+y^2)^2)) is (2x(x^2-3y^2))/((x^2+y^2)^3)

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

\frac{2 x ( x ^{2} - 3 y ^{2} )}{( x ^{2} + y ^{2} ) ^{3}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d y}{d y} = 1

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

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

Simplify

2 x \frac{1 \cdot ( x ^{2} + y ^{2} ) ^{2} - 4 y ( x ^{2} + y ^{2} ) y}{( ( x ^{2} + y ^{2} ) ^{2} ) ^{2}} : \frac{2 x ( x ^{2} - 3 y ^{2} )}{( x ^{2} + y ^{2} ) ^{3}}

= \frac{2 x ( x ^{2} - 3 y ^{2} )}{( x ^{2} + y ^{2} ) ^{3}}

\frac{d}{d x} (9 x + 1)

\int_{1}^{3} \frac{2}{( x - 2 ) ^{\frac{8}{3}}} d x

2 x^{2} - 5

\int_{- 3}^{3} x^{3} d x

\frac{d}{d x} (ln (1 + 3 x) + x ln (x))

What is the d/(dy)((2xy)/((x^2+y^2)^2)) ?
The d/(dy)((2xy)/((x^2+y^2)^2)) is (2x(x^2-3y^2))/((x^2+y^2)^3)
What is the first d/(dy)((2xy)/((x^2+y^2)^2)) ?
The first d/(dy)((2xy)/((x^2+y^2)^2)) is (2x(x^2-3y^2))/((x^2+y^2)^3)