# How do you find the derivative of ln(sqrtx)?

Dec 19, 2015

$\frac{1}{2 x}$

#### Explanation:

The chain rule can be generalized to terms in the natural logarithm as follows:

$\frac{d}{\mathrm{dx}} \left[\ln \left(u\right)\right] = \frac{u '}{u}$

Thus,

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

=(1/2x^(-1/2))/x^(1/2)=1/(2x^(1/2)x^(1/2)

$= \frac{1}{2 x}$