How do you find the second derivative of # ln(x^2+4)# ?

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Answer 1 · 2016-10-15T18:51:26

#(d^2ln(x^2 + 4))/dx^2 = (8 - 2x^2)/(x^2 + 4)^2#

#(d{f(u(x))})/dx = (df(u))/(du)((du)/dx)#

Let #u(x) = x^2 + 4#, then #(df(u))/(du) =(dln(u))/(du) = 1/u# and #(du)/dx = 2x#

#(dln(x^2 + 4))/dx = (2x)/(x^2 + 4)#

#(d^2ln(x^2 + 4))/dx^2 = (d((2x)/(x^2 + 4)))/dx#

#(d((2x)/(x^2 + 4)))/dx =#

#{2(x^2 + 4) - 2x(2x)}/(x^2 + 4)^2 = #

#(8 - 2x^2)/(x^2 + 4)^2#