Question

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

Answer 1

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)`