What is the derivative of sqrt(17-x^2-4y^2) ?

The derivative of sqrt(17-x^2-4y^2) is -(x+4y\frac{dy)/(dx)}{sqrt(-x^2-4y^2+17)}

What is the first derivative of sqrt(17-x^2-4y^2) ?

The first derivative of sqrt(17-x^2-4y^2) is -(x+4y\frac{dy)/(dx)}{sqrt(-x^2-4y^2+17)}

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\frac{d}{d x} (17 - x^{2} - 4 y^{2})

- \frac{x + 4 y \frac{d y}{d x}}{- x ^{2} - 4 y ^{2} + 17}

Solution steps

\frac{d}{d x} (17 - x^{2} - 4 y^{2})

Apply the chain rule:

\frac{1}{2 17 - x ^{2} - 4 y ^{2}} \frac{d}{d x} (17 - x^{2} - 4 y^{2})

= \frac{1}{2 17 - x ^{2} - 4 y ^{2}} \frac{d}{d x} (17 - x^{2} - 4 y^{2})

\frac{d}{d x} (17 - x^{2} - 4 y^{2}) = - 2 x - 8 y \frac{d y}{d x}

= \frac{1}{2 17 - x ^{2} - 4 y ^{2}} (- 2 x - 8 y \frac{d y}{d x})

Simplify

\frac{1}{2 17 - 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} + 17}

= - \frac{x + 4 y \frac{d y}{d x}}{- x ^{2} - 4 y ^{2} + 17}

\int (4 - x^{2})^{3} d x

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

- \frac{1}{( 3 + x ) ^{2}}

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

\int_{0}^{\infty} 9 e^{- 9 x} d x

What is the derivative of sqrt(17-x^2-4y^2) ?
The derivative of sqrt(17-x^2-4y^2) is -(x+4y\frac{dy)/(dx)}{sqrt(-x^2-4y^2+17)}
What is the first derivative of sqrt(17-x^2-4y^2) ?
The first derivative of sqrt(17-x^2-4y^2) is -(x+4y\frac{dy)/(dx)}{sqrt(-x^2-4y^2+17)}