What is (\partial)/(\partial x)((4-x^2-y^2)^{1/2}) ?

The answer to (\partial)/(\partial x)((4-x^2-y^2)^{1/2}) is -x/((4-x^2-y^2)^{1/2)}

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d t} (\frac{1}{t ^{2} - 2})

\frac{\partial}{\partial x} (ln (4 + 3 x^{2} + 2 y^{2}))

f (x) = 2 sin^{4} (x)

y^{^{''}} - 6 y^{^{'}} = - 2 x - 5

\frac{d}{d x} (\frac{3 x}{y})

What is (\partial)/(\partial x)((4-x^2-y^2)^{1/2}) ?
The answer to (\partial)/(\partial x)((4-x^2-y^2)^{1/2}) is -x/((4-x^2-y^2)^{1/2)}