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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\int 6 x - x^{2} - 8 d x

\int x sin (\frac{1}{8} x) d x

\int \frac{( x ^{3} - 6 x - 20 )}{x + 5} d x

y = (x^{4}) + (\frac{3}{( x ^{2} )}), at x = - 1

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