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 $(du)/dx = 2x$

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

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

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

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

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

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