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How do you find the first and second derivative of #lnx^(1/2)#?

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1 Answer

Shwetank Mauria

Jun 14, 2016

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

Explanation:

We use the concept of function of a function according to which,.

#d/(dx)f(g(x))=(df)/(dg)xx(dg)/(dx)#

Hence as #f(x)=lnx^(1/2)#

#f'(x)=1/x^(1/2)xx1/2x^(-1/2)=1/(2x)#

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

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