What is (\partial)/(\partial x)(sqrt(4-x^2-y^2)) ?

The answer to (\partial)/(\partial x)(sqrt(4-x^2-y^2)) is -x/(sqrt(4-x^2-y^2))

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(\partial)/(\partial x)(sqrt(4-x^2-y^2))

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

- \frac{x}{4 - x ^{2} - y ^{2}}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

Simplify

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

= - \frac{x}{4 - x ^{2} - y ^{2}}

\int \frac{3 - cos ( x )}{sin ^{2} ( x )} d x

\int_{- 8}^{0} \frac{1}{( x - 8 ) ^{2}} d x

\frac{d y}{d x} = \frac{( y ^{2} + 4 )}{2 x y}

\frac{d y}{d x} = \frac{cos ( 8 x )}{cos ( 4 y )}

\frac{s + 3}{s + 5}

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