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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Simplify

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

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

16 + t

\frac{d}{d x} ((5 x + 2)^{4} (3 x - 7))

\int e^{- x^{2}} \cdot x d x

k = 1 \sum \infty \frac{k ^{2}}{k !}

\frac{d}{d x} (\frac{x ^{3} + 2 x ^{2} - 1}{x + 5})

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