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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \cdot x + 1

3^{6}

[(2 π)^{2}]^{2}

3 x^{5} (2 x^{3} y^{3})

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