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

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

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

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

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Popular >

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (4 - x^{2} - 4 y^{2}) = - 2 x - 8 y \frac{d y}{d x}

= \frac{1}{2 4 - x ^{2} - 4 y ^{2}} (- 2 x - 8 y \frac{d y}{d x})

Simplify

\frac{1}{2 4 - x ^{2} - 4 y ^{2}} (- 2 x - 8 y \frac{d y}{d x}) : - \frac{x + 4 y \frac{d y}{d x}}{- x ^{2} - 4 y ^{2} + 4}

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

k = 1 \sum \infty e^{k}

(2, - 7), (- 1, 6)

\frac{d}{d x} (arctan (cos (x)))

\int_{7}^{14} \frac{1}{( x - 9 ) ^{2}} d x

\int_{0}^{4} ∣ 3 t - 8 ∣ d t

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