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

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

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

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

- \frac{y}{9 - x ^{2} - y ^{2}}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

\frac{1}{2 9 - x ^{2} - y ^{2}} \frac{\partial}{\partial y} (9 - x^{2} - y^{2})

= \frac{1}{2 9 - x ^{2} - y ^{2}} \frac{\partial}{\partial y} (9 - x^{2} - y^{2})

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

= \frac{1}{2 9 - x ^{2} - y ^{2}} (- 2 y)

Simplify

\frac{1}{2 9 - x ^{2} - y ^{2}} (- 2 y) : - \frac{y}{9 - x ^{2} - y ^{2}}

= - \frac{y}{9 - x ^{2} - y ^{2}}

\frac{d}{d x} (x^{9} + 9^{x})

\int x (ln (x^{2})) d x

\int \frac{- 2 z - 3}{25 ( z ^{2} + 4 )} d z

y^{^{'}} + 2 y = 5 y^{3}

y^{^{''}} - 2 y^{^{'}} + y = 2 x e^{2 x} + 6 e^{x}

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