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,...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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)).`**