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))#