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

1 Answer
Guilherme N.
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)#

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