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

Sep 4, 2015

$\frac{d}{\mathrm{dx}} \ln x \sqrt{{x}^{2} + 1} = \ln x {\left({x}^{2} + 1\right)}^{- \frac{1}{2}} + \frac{1}{x} \sqrt{{x}^{2} + 1}$
$\frac{d}{\mathrm{dx}} \ln x \sqrt{{x}^{2} + 1} = \ln x \frac{d}{\mathrm{dx}} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} + {\left({x}^{2} + 1\right)}^{\frac{1}{2}} \frac{d}{\mathrm{dx}} \ln x$
$= \left(\frac{2}{2}\right) \ln x {\left({x}^{2} + 1\right)}^{- \frac{1}{2}} + \frac{1}{x} \sqrt{{x}^{2} + 1}$