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

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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int 4 x (2 x^{2} + 1) 8 d x

\frac{d y}{d t} - (\frac{3}{t}) \cdot y = 2 t^{3} \cdot e^{(2 t)}

\frac{d}{d t} (5 e^{- 2 t} + 3 t e^{- 2 t} - 2)

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y^{^{'}} (1) = 4

x cos (x)

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