# How do you find the derivative of y=ln(sqrt(x))?

Aug 18, 2014

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

Explanation

$y = \ln \left(\sqrt{x}\right)$

differentiating with respect to $x$ using Chain Rule,

$y ' = \frac{1}{\sqrt{x}} \cdot \frac{1}{2} {x}^{- \frac{1}{2}}$

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

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

Nov 6, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 x}$

#### Explanation:

$y = \ln \left(\sqrt{x}\right) = \ln \left({x}^{\frac{1}{2}}\right) = \frac{1}{2} \ln \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 x}$