What is (\partial)/(\partial y)((x^2)/(4-y^2)) ?

The answer to (\partial)/(\partial y)((x^2)/(4-y^2)) is (2x^2y)/((4-y^2)^2)

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(\partial)/(\partial y)((x^2)/(4-y^2))

\frac{\partial}{\partial y} (\frac{x ^{2}}{4 - y ^{2}})

\frac{2 x ^{2} y}{( 4 - y ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (\frac{x ^{2}}{4 - 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}{4 - y ^{2}})

Apply exponent rule:

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

= x^{2} \frac{\partial}{\partial y} ((4 - y^{2})^{- 1})

Apply the chain rule:

- \frac{1}{( 4 - y ^{2} ) ^{2}} \frac{\partial}{\partial y} (4 - y^{2})

= - \frac{1}{( 4 - y ^{2} ) ^{2}} \frac{\partial}{\partial y} (4 - y^{2})

\frac{\partial}{\partial y} (4 - y^{2}) = - 2 y

= x^{2} (- \frac{1}{( 4 - y ^{2} ) ^{2}} (- 2 y))

Simplify

x^{2} (- \frac{1}{( 4 - y ^{2} ) ^{2}} (- 2 y)) : \frac{2 x ^{2} y}{( 4 - y ^{2} ) ^{2}}

= \frac{2 x ^{2} y}{( 4 - y ^{2} ) ^{2}}

What is (\partial)/(\partial y)((x^2)/(4-y^2)) ?
The answer to (\partial)/(\partial y)((x^2)/(4-y^2)) is (2x^2y)/((4-y^2)^2)