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

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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (I n \frac{x + 1}{x - 1})

\int x sin^{2} (2 x) d x

\frac{d}{d x} ((x^{2} + 3 x) e^{2 x})

4 t^{3}

x \to \infty + lim (4 x)

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