# How do I find the derivative of  (ln x)^(1/2)?

Jan 9, 2016

$\frac{1}{2 x \sqrt{\ln} x}$

#### Explanation:

Use the chain rule here:

$\frac{d}{\mathrm{dx}} \left({u}^{\frac{1}{2}}\right) = \frac{1}{2} {u}^{- \frac{1}{2}} \cdot u ' = \frac{1}{2 \sqrt{u}} \cdot u '$

Thus, when $u = \ln x$,

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

Since $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$, the derivative of the original function is

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

or, if you prefer fractional exponents

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

Jan 9, 2016

We'll need chain rule to solve this one.

#### Explanation:

• Chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

In this case, we'll make the function differentiable by renaming $u = \ln x$, so that $f \left(x\right) = {u}^{\frac{1}{2}}$.

Now, let's proceed following chain rule statement:

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