# What is the derivative of sqrt(x^2+1)?

Mar 12, 2016

$\frac{x}{\sqrt{{x}^{2} + 1}}$

#### Explanation:

Using the chain rule
If a function $f \left(x\right) = {\left[g \left(x\right)\right]}^{n}$

Then $f ' \left(x\right) = n {\left[g \left(x\right)\right]}^{n - 1} \cdot g ' \left(x\right)$

Given $g \left(x\right) = \left({x}^{2} + 1\right) \mathmr{and} n = \frac{1}{2}$
We see that $g ' \left(x\right) = 2 x$
$\therefore \textrm{\mathrm{de} r i v a t i v e o f} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} = \frac{1}{2} {\left[{x}^{2} + 1\right]}^{\frac{1}{2} - 1} \cdot 2 x$
$= \frac{1}{\cancel{2}} {\left[{x}^{2} + 1\right]}^{- \frac{1}{2}} \cdot \cancel{2} x$
or $= x {\left[{x}^{2} + 1\right]}^{- \frac{1}{2}}$
or $= \frac{x}{\sqrt{{x}^{2} + 1}}$