What is the second derivative of #f(x)= ln sqrt(x)#?

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Answer 1 · 2016-02-28T14:05:36

The second derivative is

#d^2(lnsqrtx)/dx^2=d/dx(dlnsqrtx/dx)=d/dx(1/2*1/x)= =-1/2*1/x^2#

Note that #lnsqrtx=lnx^(1/2)=1/2*lnx#

Answer 2 · 2016-02-28T14:32:05

#f''(x)=-1/(2x^2)#

We can simplify #f(x)# through the rule that

#log_a(b^c)=c*log_a(b)#

Here, we see that

#f(x)=lnsqrtx=ln(x^(1/2))=1/2ln(x)#

So, to find the derivative of this function, we must know that the derivative of #ln(x)# is #"1/"x#, so

#f'(x)=1/2(1/x)#

To differentiate this and find the second derivative, use the power rule, recalling that #"1/"x=x^-1#.

#f''(x)=1/2d/dx(x^-1)=1/2(-x^-2)=-1/(2x^2)#