# How do you find the derivative of sqrt(x^2+1)?

Apr 14, 2015

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

This is a power and the chain rule. (A combination sometimes called "The General Power Rule: $\frac{d}{\mathrm{dx}} \left({\left(g \left(x\right)\right)}^{r}\right) = r {\left(g \left(x\right)\right)}^{r - 1} g ' \left(x\right)$)

Because the square root appears so often, I encourage students to memorize:

$\frac{d}{\mathrm{dx}} \sqrt{x} = \frac{1}{2 \sqrt{x}}$

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