# Derivative of square root of x^2+y^2 | Derivative of x/sqrt{x^2+y^2}

In this post, we will learn how to find the derivative of various square roots; for example, the derivatives of root(x^2+y^2), x/root(x^2+y^2) and y/root(x^2+y^2).

For more details of square roots, please visit the page Square Root of x: Definition, Symbol, Graph, Properties, Derivative, Integration.

## Derivative of $\sqrt{x^2+y^2}$

To find the derivative of $\sqrt{x^2+y^2}$ with respect to $x$, we will apply the following rule (it is obtained by the chain rule of derivatives):

$\dfrac{d}{dx}(\sqrt u) = \dfrac{1}{2\sqrt{u}} \dfrac{du}{dx}$ $\cdots (i)$

Let $u:=u(x)$ be a function of $x$. Then the derivative of the square root of $u$ with respect to $x$ is given as follows:

Step 1:

Let $u=x^2+y^2$

At first, we will apply the above rule (i). By doing so we get that

$\therefore \dfrac{d}{dx}(\sqrt{x^2+y^2})$ $=\dfrac{1}{2\sqrt{x^2+y^2}} * \dfrac{d}{dx}(\sqrt{x^2+y^2})$ $\cdots (ii)$

Step 2:

As we know that $\dfrac{d}{dx}(x^2)=2x$ and $\dfrac{d}{dx}(y^2)=2y \dfrac{dy}{dx}$, we obtain from (ii) that

$\dfrac{d}{dx}(\sqrt{x^2+y^2})$ $=\dfrac{1}{2\sqrt{x^2+y^2}} * \dfrac{d}{dx}(\sqrt{x^2+y^2})$

$=\dfrac{1}{2\sqrt{x^2+y^2}} (2x+2y \dfrac{dy}{dx})$

$=\dfrac{2}{2\sqrt{x^2+y^2}} (x+y \dfrac{dy}{dx})$

$=\dfrac{1}{\sqrt{x^2+y^2}} (x+y \dfrac{dy}{dx})$

Thus, the derivative of $\sqrt{x^2+y^2}$ is equal to

$\dfrac{d}{dx}(\sqrt{x^2+y^2})$ $=\dfrac{1}{\sqrt{x^2+y^2}}(x+y \dfrac{dy}{dx})$ $=\dfrac{x}{\sqrt{x^2+y^2}}$ $+\dfrac{y}{\sqrt{x^2+y^2}} \dfrac{dy}{dx}$ $\cdots (\star)$ Answer.

## Derivative of  $\dfrac{x}{\sqrt{x^2+y^2}}$

By the quotient rule of derivatives, the derivative of $\dfrac{x}{\sqrt{x^2+y^2}}$ is equal to

$=\dfrac{\sqrt{x^2+y^2} \frac{dx}{dx} – x \dfrac{d}{dx}(\sqrt{x^2+y^2})}{(\sqrt{x^2+y^2})^2}$

$=\dfrac{1}{x^2+y^2} \times$ $\sqrt{x^2+y^2}$ $-x[\dfrac{1}{\sqrt{x^2+y^2}}(x+y \dfrac{dy}{dx})]$ [from $(\star)$]

$=\dfrac{1}{(x^2+y^2)^{3/2}} \times$ $(x^2+y^2-x^2-xy \dfrac{dy}{dx})$

$=\dfrac{1}{(x^2+y^2)^{3/2}}(y^2-xy \dfrac{dy}{dx})$

$=\dfrac{y}{(x^2+y^2)^{3/2}}(y-x \dfrac{dy}{dx})$

## Derivative of  $\dfrac{y}{\sqrt{x^2+y^2}}$

By the quotient rule of derivatives, the derivative of $\dfrac{y}{\sqrt{x^2+y^2}}$ is equal to

$=\dfrac{\sqrt{x^2+y^2} \dfrac{dy}{dx} – y \dfrac{d}{dx}(\sqrt{x^2+y^2})}{(\sqrt{x^2+y^2})^2}$

$=\dfrac{1}{x^2+y^2} \times$ $\sqrt{x^2+y^2} \dfrac{dy}{dx}$ $-y[\dfrac{1}{\sqrt{x^2+y^2}}(x+y \dfrac{dy}{dx})]$ [from $(\star)$]

$=\dfrac{1}{(x^2+y^2)^{3/2}} \times$ $((x^2+y^2) \dfrac{dy}{dx}-xy-y^2 \dfrac{dy}{dx})$

$=\dfrac{1}{(x^2+y^2)^{3/2}}((x^2+y^2-y^2)\dfrac{dy}{dx}-xy)$

$=\dfrac{1}{(x^2+y^2)^{3/2}}(x^2 \dfrac{dy}{dx}-xy)$

$=\dfrac{x}{(x^2+y^2)^{3/2}}(x \dfrac{dy}{dx} – y)$