Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 23

Find y' and y''.
For the first derivative, use the product rule AB'+BA'.
Be warned that since we have a ln(2x) term, we will need to use chain rule, which is the derivative of the inner term 2x.
y' = x^2* (1/(2x)) * 2 + ln(2x) *2x
Simplify the above equation.
The first derivative is: y'=x + 2xln(2x)
To find the second derivative, we will need to take the derivative of each term. The derivative of first term is simply 1. The second term will require product rule and chain rule like what we have done for the first derivative.
y'=x+[(2x)(ln(2x))]
Apply the derivative.
y'' = 1 + [(2x)(1/ (2x))*2 + ln(2x) (2)]
Simplify.
y''= 1+2+2ln(2x)
The second derivative is: y''= 2ln(2x)+3