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

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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

= 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

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

Simplify

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

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

n = 1 \sum \infty \frac{( - 1 ) ^{n - 1} n ^{2}}{1 2 ^{n}}

\frac{s + 3}{( s + 3 ) ^{2} + 9}

\frac{d}{d x} (x ln (- x))

x \to π lim (\frac{tan ( x )}{x})

\int_{7}^{\infty} x e^{- 5 x} d x

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