What is (\partial)/(\partial y)(9-x^2-y^2) ?

The answer to (\partial)/(\partial y)(9-x^2-y^2) is -2y

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(\partial)/(\partial y)(9-x^2-y^2)

\frac{\partial}{\partial y} (9 - x^{2} - y^{2})

- 2 y

Solution steps

\frac{\partial}{\partial y} (9 - x^{2} - y^{2})

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (9) - \frac{\partial}{\partial y} (x^{2}) - \frac{\partial}{\partial y} (y^{2})

\frac{\partial}{\partial y} (9) = 0

\frac{\partial}{\partial y} (x^{2}) = 0

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

= 0 - 0 - 2 y

Simplify

= - 2 y

x \to \infty lim (\frac{10 + 9 e ^{(2 x)}}{11 - 8 e ^{(3 x)}})

\frac{d}{d x} (a r (x) cos (x))

\int_{0}^{5} x d x

ln (9 sin^{2} (x))

\frac{d}{d u} (\frac{5 u}{4 u ^{2} + 1})

What is (\partial)/(\partial y)(9-x^2-y^2) ?
The answer to (\partial)/(\partial y)(9-x^2-y^2) is -2y