Find the first and second derivative of #(2lnx)/x#?
1 Answer
Apr 10, 2017
# f'(x) \ = (2lnx)/x #
# f''(x) = (2-2lnx)/(x^2)#
Explanation:
We have:
# f(x) = ln^2x #
Differentiating once (using the chain rule) we get:
# f'(x) = (2lnx )(d/d ln x )#
# " " = (2lnx )(1/x) #
# " " = (2lnx)/x #
Ti get the second derivative we apply the quotient rule:
# f''(x) = ((x)(d/dx2lnx)-(2lnx)(d/dxx))/(x^2)#
# " " = ((x)(2/x)-(2lnx)(1))/(x^2)#
# " " = (2-2lnx)/(x^2)#
