What is (\partial)/(\partial x)(sqrt(3-x^2-y^2)) ?

The answer to (\partial)/(\partial x)(sqrt(3-x^2-y^2)) is -x/(sqrt(3-x^2-y^2))

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(\partial)/(\partial x)(sqrt(3-x^2-y^2))

\frac{\partial}{\partial x} (3 - x^{2} - y^{2})

- \frac{x}{3 - x ^{2} - y ^{2}}

Solution steps

\frac{\partial}{\partial x} (3 - x^{2} - y^{2})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 3 - x ^{2} - y ^{2}} \frac{\partial}{\partial x} (3 - x^{2} - y^{2})

= \frac{1}{2 3 - x ^{2} - y ^{2}} \frac{\partial}{\partial x} (3 - x^{2} - y^{2})

\frac{\partial}{\partial x} (3 - x^{2} - y^{2}) = - 2 x

= \frac{1}{2 3 - x ^{2} - y ^{2}} (- 2 x)

Simplify

\frac{1}{2 3 - x ^{2} - y ^{2}} (- 2 x) : - \frac{x}{3 - x ^{2} - y ^{2}}

= - \frac{x}{3 - x ^{2} - y ^{2}}

\frac{d}{d x} (x \cdot x^{2} - 1)

\frac{d}{d x} (\frac{1}{lo g _{10} ( x )})

\frac{\partial}{\partial y} (4 x^{3} y^{3} - 1)

\int \frac{4 + x + x ^{2}}{x} d x

y^{^{'}} = - \frac{y}{100}, y (0) = 80

What is (\partial)/(\partial x)(sqrt(3-x^2-y^2)) ?
The answer to (\partial)/(\partial x)(sqrt(3-x^2-y^2)) is -x/(sqrt(3-x^2-y^2))