# How do you find the derivative of f(x)=sqrt(a^2+x^2)?

Sep 17, 2016

$f ' \left(x\right) = \frac{x}{\sqrt{{a}^{2} + {x}^{2}}}$

#### Explanation:

The chain rule goes like this:
If $f \left(x\right) = {\left(g \left(x\right)\right)}^{n}$, then $f ' \left(x\right) = n {\left(g \left(x\right)\right)}^{n - 1} \cdot \frac{d}{\mathrm{dx}} g \left(x\right)$

Applying this rule:

$f \left(x\right) = \sqrt{{a}^{2} + {x}^{2}} = {\left({a}^{2} + {x}^{2}\right)}^{\frac{1}{2}}$

$f ' \left(x\right) = \frac{1}{2} {\left({a}^{2} + {x}^{2}\right)}^{\frac{1}{2} - 1} \cdot \frac{d}{\mathrm{dx}} \left({a}^{2} + {x}^{2}\right)$

$f ' \left(x\right) = \frac{1}{2} {\left({a}^{2} + {x}^{2}\right)}^{- \frac{1}{2}} \cdot 2 x$

$f ' \left(x\right) = \frac{1}{2 {\left({a}^{2} + {x}^{2}\right)}^{\frac{1}{2}}} \cdot 2 x$

$f ' \left(x\right) = \frac{x}{{\left({a}^{2} + {x}^{2}\right)}^{\frac{1}{2}}}$

$f ' \left(x\right) = \frac{x}{\sqrt{{a}^{2} + {x}^{2}}}$