What is the second derivative of f(x)= ln sqrt(xe^x)?

1 Answer
Jan 6, 2016

You'll need product rule and chain rule, as well as acknowledging that the derivative of lnx=1/x.

Explanation:

  • Product rule: (ab)'=a'b+ab'
  • Chain rule: (dy)/(dx)=(dy)/(dv)(dv)/(du)(du)/(dx)

renaming u=xe^x and v=sqrt(u), we can proceed:

(dy)/(dx)=1/v1/(2sqrt(u))(e^x+xe^x)=(e^x(1+x))/(2vsqrt(u))

Substituting v, then u:

(dy)/(dx)=(e^x(1+x))/(2sqrt(u)sqrt(u))=(e^x(1+x))/(2u)=(cancel(e^x)(1+x))/(2xcancel(e^x))

(dy)/(dx)=(1+x)/(2x)

The second derivative will demand quotient rule: (a/b)'=(a'b-ab')/b^2

(dy^2)/(d^2x)=(1(2x)-(1+x)2)/(4x^2)=(2x-2-2x)/(4x^2)=-2/(4x^2)=-1/(2x^2)