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

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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d P}{d t} = P + a

5 - 4 x, \frac{1}{x}

\frac{3 x - 8}{7 x + 4}

y^{^{'}} - \frac{4}{x} y = \frac{y ^{3}}{x ^{5}}, y (1) = 1

(1.1) (2.5)

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