Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 25

You need to find the first derivative of the function, using the chain rule, such that:
y' = (ln(x+sqrt(1+x^2)))'
y' = (ln'(x+sqrt(1+x^2)))*(x+sqrt(1+x^2))'
y' = (1/(x+sqrt(1+x^2)))*(1+(2x)/(2sqrt(1+x^2)))
y' = (1/(x+sqrt(1+x^2)))*(1+(x)/(sqrt(1+x^2)))
y' = (sqrt(1+x^2)) + x)/((sqrt(1+x^2))(x+sqrt(1+x^2)))
Reducing like terms, yields:
y' = 1/(sqrt(1+x^2))
You need to evaluate the second derivative, differentiating the first derivative, with respect to x,using the quotient and chain rules, such that:
y'' = ((1)'(sqrt(1+x^2)) - 1*(sqrt(1+x^2))')/(1+x^2)
y'' = -(x)/((1+x^2)^(3/2))
Hence, evaluating the first and the second derivatives, yields y' = 1/(sqrt(1+x^2)) and y'' = -(x)/((1+x^2)^(3/2)).