# Step-by-step Solution

## Find the derivative using the quotient rule (d/dx)(x/((x^2+y^2)^0.5))

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$\frac{\sqrt{x^2+y^2}-\left(x^2+y^2\right)^{-\frac{1}{2}}x^2}{x^2+y^2}$

## Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\frac{x}{\sqrt{x^2+y^2}}\right)$
1

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\sqrt{x^2+y^2}\cdot\frac{d}{dx}\left(x\right)-x\frac{d}{dx}\left(\sqrt{x^2+y^2}\right)}{\left(\sqrt{x^2+y^2}\right)^2}$
2

The derivative of the linear function is equal to $1$

$\frac{\sqrt{x^2+y^2}-x\frac{d}{dx}\left(\sqrt{x^2+y^2}\right)}{\left(\sqrt{x^2+y^2}\right)^2}$

$\frac{\sqrt{x^2+y^2}-\left(x^2+y^2\right)^{-\frac{1}{2}}x^2}{x^2+y^2}$

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$\frac{d}{dx}\left(\frac{x}{\sqrt{x^2+y^2}}\right)$

### Main topic:

Differential calculus

~ 0.52 seconds