How do you find the second derivative of ln(x^2+4)ln(x2+4) ?

1 Answer
Oct 15, 2016

(d^2ln(x^2 + 4))/dx^2 = (8 - 2x^2)/(x^2 + 4)^2d2ln(x2+4)dx2=82x2(x2+4)2

Explanation:

The chain rule is:

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

Let u(x) = x^2 + 4u(x)=x2+4, then (df(u))/(du) =(dln(u))/(du) = 1/udf(u)du=dln(u)du=1u and (du)/dx = 2xdudx=2x

(dln(x^2 + 4))/dx = (2x)/(x^2 + 4)dln(x2+4)dx=2xx2+4

(d^2ln(x^2 + 4))/dx^2 = (d((2x)/(x^2 + 4)))/dxd2ln(x2+4)dx2=d(2xx2+4)dx

(d((2x)/(x^2 + 4)))/dx =d(2xx2+4)dx=

{2(x^2 + 4) - 2x(2x)}/(x^2 + 4)^2 = 2(x2+4)2x(2x)(x2+4)2=

(8 - 2x^2)/(x^2 + 4)^282x2(x2+4)2