`h(x) = ln(x + sqrt(x^2 - 1))` Differentiate the function.

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Chapter 3, 3.6 - Problem 12 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sbernabel | Student, Graduate | (Level 1) Adjunct Educator

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In order to find h'(x) we need to know a few things:

The derivative of log(f(x)), and of sqrt(x^2-1).

We need to know the Chain rule. Given h(x)=log(x+sqrt(x^2-1)), we start by differentiating log(f(x)), where f(x) = x+sqrt(x^2-1):

d/dx (log(f(x))) = 1/f(x) * d/dx(f(x)).

Now find the derivative of f(x), that is

d/dx (x+sqrt(g(x))), where g(x) = x^2-1

We have d/dx(x+sqrt(g(x))) = 1 + 1/(2*sqrt(g(x)) * d/dx(g(x))

Finally we are left with finding the derivative of g(x), simple enough.

d/dx (x^2-1) = 2*x

Back substituting we have

d/dx (h(x)) = 1/(x+sqrt(x^2-1)) * (1 + 1/(2*sqrt(x^2-1))*2*x)

Simplifying, h'(x) = 1/(x+sqrt(x^2-1)) * (1 + x/(sqrt(x^2-1))

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