What is the derivative of sqrt(4-x^2-y^2) ?

The derivative of sqrt(4-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+4)}

What is the first derivative of sqrt(4-x^2-y^2) ?

The first derivative of sqrt(4-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+4)}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

lo g_{10} (x + 5)

x (\frac{d y}{d x}) + 3 y = \frac{1}{x}

\frac{d}{d x} (sin (2 x) sin (x))

\frac{d y}{d x}, 25 x = y^{4}

\frac{d}{d x} ((9 x^{6} + 4 x^{3})^{4})

What is the derivative of sqrt(4-x^2-y^2) ?
The derivative of sqrt(4-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+4)}
What is the first derivative of sqrt(4-x^2-y^2) ?
The first derivative of sqrt(4-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+4)}