What is the derivative of sqrt(x^2+y^2-1) ?

The derivative of sqrt(x^2+y^2-1) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-1)}

What is the first derivative of sqrt(x^2+y^2-1) ?

The first derivative of sqrt(x^2+y^2-1) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-1)}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{1}{3 x ^{2} + 6 x - 4} d x

n = 1 \sum \infty \frac{1}{n + 5}

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

\frac{d y}{d x} + 2 y = 3 x

y^{^{''}} - 10 y^{^{'}} + 26 y = 0

What is the derivative of sqrt(x^2+y^2-1) ?
The derivative of sqrt(x^2+y^2-1) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-1)}
What is the first derivative of sqrt(x^2+y^2-1) ?
The first derivative of sqrt(x^2+y^2-1) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-1)}