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

The d/(dy)((2x)/(x^2+y^2)) is -(4xy)/((x^2+y^2)^2)

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

The first d/(dy)((2x)/(x^2+y^2)) is -(4xy)/((x^2+y^2)^2)

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d 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}}

\frac{d}{d x} ((4 - x^{2})^{3})

\int \frac{sin ( \frac{2}{x} )}{x ^{2}} d x

\frac{d}{d x} (\frac{5}{3} x^{2} y^{- 2})

\int \frac{2}{x ^{2} + 9} d x

n = 0 \sum \infty \frac{n}{1 + n ^{2}}

What is the d/(dy)((2x)/(x^2+y^2)) ?
The d/(dy)((2x)/(x^2+y^2)) is -(4xy)/((x^2+y^2)^2)
What is the first d/(dy)((2x)/(x^2+y^2)) ?
The first d/(dy)((2x)/(x^2+y^2)) is -(4xy)/((x^2+y^2)^2)