What is (\partial)/(\partial y)(x/(1+x^2+y^2)) ?

The answer to (\partial)/(\partial y)(x/(1+x^2+y^2)) is -(2xy)/((1+x^2+y^2)^2)

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(\partial)/(\partial y)(x/(1+x^2+y^2))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{0}^{1} \frac{9 x}{e ^{2 x}} d x

\frac{d}{d x} (1 + \frac{3}{x})

\int \frac{1}{50 + t} d t

f (x) = 3 t^{4} - 7 t^{3} + 2

What is (\partial)/(\partial y)(x/(1+x^2+y^2)) ?
The answer to (\partial)/(\partial y)(x/(1+x^2+y^2)) is -(2xy)/((1+x^2+y^2)^2)