# What is the derivative of ln(sqrtx)?

Oct 19, 2015

As an alternative, use $\ln \sqrt{x} = \ln \left({x}^{\frac{1}{2}}\right) = \frac{1}{2} \ln x$

#### Explanation:

So,

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

Oct 19, 2015

More detailed explanation!

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{1}{2 x}$

#### Explanation:

Pre amble:
Suppose you have$\ln \left(z\right) = s$. This may be rewritten as $z = {e}^{s}$.
You also need to know that if we have $\frac{d}{\mathrm{ds}} \left({e}^{s}\right)$ then the solution is $\frac{d}{\mathrm{ds}} \left({e}^{s}\right) = {e}^{s}$

Back to the question:
$y = \ln \left(\sqrt{x}\right) = \ln \left({x}^{\frac{1}{2}}\right)$

So ${x}^{\frac{1}{2}} = {e}^{y}$ or alternatively squaring both sides gives:

$x = {\left({e}^{y}\right)}^{2}$ ...................................(1)

The differential of this is:
$\frac{\mathrm{dx}}{\mathrm{dy}} = 2 \left({e}^{y}\right)$ ...............................(2)

But we need $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$ this can be achieved by inverting (2) giving

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left[2 \left({e}^{y}\right)\right]}^{- 1} = \frac{1}{2 \left({e}^{y}\right)}$................(3)

But from (1) we have ${e}^{y} = x$ so by substation in (3) we have:

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{1}{2 x}$