`y = ln(x + sqrt(1 + x^2))` Find y' and y''.

Textbook Question

Chapter 3, 3.6 - Problem 25 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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)).`

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