What is the derivative of sqrt(16-x^2-y^2) ?

The derivative of sqrt(16-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+16)}

What is the first derivative of sqrt(16-x^2-y^2) ?

The first derivative of sqrt(16-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+16)}

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\frac{d}{d x} (16 - x^{2} - y^{2})

- \frac{x + y \frac{d y}{d x}}{- x ^{2} - y ^{2} + 16}

Solution steps

\frac{d}{d x} (16 - x^{2} - y^{2})

Apply the chain rule:

\frac{1}{2 16 - x ^{2} - y ^{2}} \frac{d}{d x} (16 - x^{2} - y^{2})

= \frac{1}{2 16 - x ^{2} - y ^{2}} \frac{d}{d x} (16 - x^{2} - y^{2})

\frac{d}{d x} (16 - x^{2} - y^{2}) = - 2 x - 2 y \frac{d y}{d x}

= \frac{1}{2 16 - x ^{2} - y ^{2}} (- 2 x - 2 y \frac{d y}{d x})

Simplify

\frac{1}{2 16 - 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} + 16}

= - \frac{x + y \frac{d y}{d x}}{- x ^{2} - y ^{2} + 16}

\int \frac{x ^{3}}{x ^{2} - 36} d x

\frac{\partial}{\partial x} (e^{2 x} \cdot cos (x + y))

y = ∣ 2 x ∣, y = x^{2} - 3

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

\frac{d}{d x} (- 8 sin (4 x))

What is the derivative of sqrt(16-x^2-y^2) ?
The derivative of sqrt(16-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+16)}
What is the first derivative of sqrt(16-x^2-y^2) ?
The first derivative of sqrt(16-x^2-y^2) is -(x+y\frac{dy)/(dx)}{sqrt(-x^2-y^2+16)}