# Question fe9da

Jan 17, 2018

Note that we have several functions in $y = x {\left(\ln x\right)}^{\frac{1}{2}}$: there is the function $f \left(x\right) = x$, which multiplies the composite function $g \left(x\right) = {\left(\ln x\right)}^{\frac{1}{2}}$. $g \left(x\right)$ is nothing more thant the composite function of $h \left(x\right) = {x}^{\frac{1}{2}}$ and $j \left(x\right) = \ln \left(x\right)$. Therefore, to find the derivative of $y$, we will need to use the product and chain rule.

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

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

Since $\frac{d}{\mathrm{dx}} \left(x\right) = 1$, then:

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

Now, the derivative $\frac{d}{\mathrm{dx}} \left[{\left(\ln x\right)}^{\frac{1}{2}}\right]$ is where we apply the chain rule. If we take $u = \ln \left(x\right)$, then:

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

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

Now we return to our original variable, $x$, since we know the relation between $u$ and $x$:

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

Then:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\ln x\right)}^{\frac{1}{2}} + x \frac{.1}{2} {\left[\ln \left(x\right)\right]}^{- \frac{1}{2}} . \left(\frac{1}{x}\right)$;

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\ln x\right)}^{\frac{1}{2}} + \frac{1}{2} {\left[\ln \left(x\right)\right]}^{- \frac{1}{2}}$.

If you want to, you can also rewrite this expression:

dy/(dx) = sqrt(ln(x)) + 1/(2sqrt(ln(x)).
dy/(dx) = (2ln(x) + 1)/(2sqrt(ln(x))#.