What is the d/(dy)(x/(x^2+y^2+1)) ?

The d/(dy)(x/(x^2+y^2+1)) is -(2xy)/((x^2+y^2+1)^2)

What is the first d/(dy)(x/(x^2+y^2+1)) ?

The first d/(dy)(x/(x^2+y^2+1)) is -(2xy)/((x^2+y^2+1)^2)

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

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

Solution steps

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

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} + 1})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to - 2 lim (\frac{5 x - 4}{4 - x})

x \to 0 lim (\frac{1 - cos ( 4 x )}{x})

\int 6 y - x - 4 d y

x \to 2 lim (1 - x)

\int_{0}^{2} 6 x^{3} x^{2} + 4 d x

What is the d/(dy)(x/(x^2+y^2+1)) ?
The d/(dy)(x/(x^2+y^2+1)) is -(2xy)/((x^2+y^2+1)^2)
What is the first d/(dy)(x/(x^2+y^2+1)) ?
The first d/(dy)(x/(x^2+y^2+1)) is -(2xy)/((x^2+y^2+1)^2)