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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to \frac{π}{2} + lim (\frac{9}{x} sec (x))

\frac{\partial}{\partial y} (e^{z} cos (y))

f (x) = \frac{10}{1 + e ^{- x}}, at x = 0

n = 2 \sum \infty (\frac{n + 1}{n})^{n}

\int \frac{5 ^{x}}{1 + 5 ^{x}} d x

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