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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

(y - x) d x + 4 x d y = 0

\int \frac{1}{2 x - 1} d x

ln ∣ sin (x) ∣

\int x^{4} (2 x^{5} - 5)^{4} d x

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

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