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Answer 1 · 2018-01-17T19:41:45

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

$\frac{d y}{d x} = \frac{d}{d x} [x {(ln x)}^{\frac{1}{2}}]$ $dy/(dx) = d/(dx)[x(lnx)^(1/2)]$ ;

$\frac{d y}{d x} = \frac{d}{d x} (x) . {(ln x)}^{\frac{1}{2}} + x . \frac{d}{d x} [{(ln x)}^{\frac{1}{2}}] .$ $dy/(dx) = d/(dx)(x) . (lnx)^(1/2) + x.d/dx[(lnx)^(1/2)].$

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

$\frac{d y}{d x} = {(ln x)}^{\frac{1}{2}} + x . \frac{d}{d x} [{(ln x)}^{\frac{1}{2}}] .$ $dy/(dx) = (lnx)^(1/2) + x.d/dx[(lnx)^(1/2)].$

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

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

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

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

$\frac{d}{d x} [{(ln x)}^{\frac{1}{2}}] = \frac{1}{2} {[ln (x)]}^{- \frac{1}{2}} . (\frac{1}{x})$ $d/dx[(lnx)^(1/2)] = 1/2 [ln(x)]^(-1/2).(1/x)$ .

Then:

$\frac{d y}{d x} = {(ln x)}^{\frac{1}{2}} + x \frac{.1}{2} {[ln (x)]}^{- \frac{1}{2}} . (\frac{1}{x})$ $dy/(dx) = (lnx)^(1/2) + x.1/2 [ln(x)]^(-1/2).(1/x)$ ;

$\frac{d y}{d x} = {(ln x)}^{\frac{1}{2}} + \frac{1}{2} {[ln (x)]}^{- \frac{1}{2}}$ $dy/(dx) = (lnx)^(1/2) + 1/2 [ln(x)]^(-1/2)$ .

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

$\frac{d y}{d x} = \sqrt{ln (x)} + \frac{1}{2 \sqrt{ln (x)}}$ $dy/(dx) = sqrt(ln(x)) + 1/(2sqrt(ln(x))$ .
$\frac{d y}{d x} = \frac{2 ln (x) + 1}{2 \sqrt{ln (x)}}$ $dy/(dx) = (2ln(x) + 1)/(2sqrt(ln(x))$ .

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