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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (- \frac{a}{x} + \frac{b}{x ^{2}})

- sin (2 x)

n = 0 \sum \infty 6

x \to \infty lim (π)

x = 4 y^{3}, x = 4 y^{2}

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