What is the derivative of sqrt(1-x^2-y^2) ?

The derivative of sqrt(1-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+1)}

What is the first derivative of sqrt(1-x^2-y^2) ?

The first derivative of sqrt(1-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+1)}

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

- \frac{x + y \frac{d y}{d x}}{- x ^{2} - y ^{2} + 1}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{2 1 - 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} + 1}

= - \frac{x + y \frac{d y}{d x}}{- x ^{2} - y ^{2} + 1}

\int_{0}^{8} y + 1 d y

x \to 0 + lim (tan (\frac{4}{x}))

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

t (\frac{d x}{d t}) + 3 x = 5 t^{2}

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

What is the derivative of sqrt(1-x^2-y^2) ?
The derivative of sqrt(1-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+1)}
What is the first derivative of sqrt(1-x^2-y^2) ?
The first derivative of sqrt(1-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+1)}