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

Sep 4, 2015

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

#### Explanation:

Since this is a product of 2 functions, we use the product rule of differentiation which states that the derivative of a product of 2 functions is the first function times the derivative of the second, plus the second function times the derivative of the first.

To differentiate the first function, ln x, we use the rule saying its 1/x, and to differentiate the second function, we write the root as a power such as to use the power rule for differentiation which states that the derivative of a function raised to a power is the function raised to one less than the given power, multiplied by the derivative of the function, which in this case is a polynomial.

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