What is (\partial)/(\partial x)((x^2)/(16-y^2)) ?

The answer to (\partial)/(\partial x)((x^2)/(16-y^2)) is (2x)/(16-y^2)

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(\partial)/(\partial x)((x^2)/(16-y^2))

\frac{\partial}{\partial x} (\frac{x ^{2}}{16 - y ^{2}})

\frac{2 x}{16 - y ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{x ^{2}}{16 - y ^{2}})

Treat

y

as a constant

Take the constant out:

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

= \frac{1}{16 - y ^{2}} \frac{\partial}{\partial x} (x^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= \frac{1}{16 - y ^{2}} \cdot 2 x^{2 - 1}

Simplify

\frac{1}{16 - y ^{2}} \cdot 2 x^{2 - 1} : \frac{2 x}{16 - y ^{2}}

= \frac{2 x}{16 - y ^{2}}

y^{'}, x^{2} - y^{2} = 9

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What is (\partial)/(\partial x)((x^2)/(16-y^2)) ?
The answer to (\partial)/(\partial x)((x^2)/(16-y^2)) is (2x)/(16-y^2)