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

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Answer 1 · 2016-01-09T18:12:11

$1/(2xsqrtlnx)$

Use the chain rule here:

$d/dx(u^(1/2))=1/2u^(-1/2)*u'=1/(2sqrtu)*u'$

Thus, when $u=lnx$ ,

$d/dx((lnx)^(1/2))=1/(2sqrtlnx)*d/dx(lnx)$

Since $d/dx(lnx)=1/x$ , the derivative of the original function is

$=1/(2xsqrtlnx)$

or, if you prefer fractional exponents

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

Answer 2 · 2016-01-09T18:13:32

We'll need chain rule to solve this one.

In this case, we'll make the function differentiable by renaming $u=lnx$ , so that $f(x)=u^(1/2)$ .

Now, let's proceed following chain rule statement:

$(dy)/(dx)=1/(2u^(1/2))(1/x)=1/(2x(lnx)^(1/2))$

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