Find the Derivative f(y) = square root of x^2+y^2

$f (y) = \sqrt{x^{2} + y^{2}}$

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + y^{2}}$ as ${(x^{2} + y^{2})}^{\frac{1}{2}}$ .

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

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x^{2} + y^{2}$ .

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To apply the Chain Rule, set $u$ as $x^{2} + y^{2}$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x^{2} + y^{2}]$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{2}$ .

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} + y^{2}]$

Replace all occurrences of $u$ with $x^{2} + y^{2}$ .

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} {(x^{2} + y^{2})}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [x^{2} + y^{2}]$

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{1}{2} {(x^{2} + y^{2})}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [x^{2} + y^{2}]$

Combine the numerators over the common denominator.

$\frac{1}{2} {(x^{2} + y^{2})}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [x^{2} + y^{2}]$

Simplify the numerator.

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Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Combine fractions.

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Move the negative in front of the fraction.

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

Combine $\frac{1}{2}$ and ${(x^{2} + y^{2})}^{- \frac{1}{2}}$ .

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

Move ${(x^{2} + y^{2})}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

By the Sum Rule, the derivative of $x^{2} + y^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [y^{2}]$ .

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

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Since $y^{2}$ is constant with respect to $x$ , the derivative of $y^{2}$ with respect to $x$ is $0$ .

$\frac{1}{2 {(x^{2} + y^{2})}^{\frac{1}{2}}} (2 x + 0)$

Simplify terms.

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Add $2 x$ and $0$ .

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

Combine $2$ and $\frac{1}{2 {(x^{2} + y^{2})}^{\frac{1}{2}}}$ .

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

Combine $\frac{2}{2 {(x^{2} + y^{2})}^{\frac{1}{2}}}$ and $x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Find the Derivative f(y) = square root of x^2+y^2

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