Calculus of a Single Variable, Chapter 5, 5.1, Section 5.1, Problem 48

y=ln(sqrt(x^2-4))
First, use the formula:
(lnu)'= 1/u*u'
Applying that formula, the derivative of the function will be:
y' =1/sqrt(x^2-4) * (sqrt(x^2-4))'
To take the derivative of the inner function, express the radical in exponent form.
y'=1/sqrt(x^2-4)*((x^2-4)^(1/2))'
Then, use the formula:
(u^n)'=n*u^(n-1) * u'
So, y' will become:
y'=1/sqrt(x^2-4) * 1/2(x^2-4)^(-1/2)*(x^2-4)'
To take the derivative of the innermost function, use the formulas:
(x^n)'=n*x^(n-1)
(c)' = 0
Applying these two formulas, y' will become:
y'=1/sqrt(x^2-4) *1/2(x^2-4)^(-1/2)*(2x-0)
Simplifying it will result to:
y'=1/sqrt(x^2-4)*1/2(x^2-4)^(-1/2)*2x
y'=1/sqrt(x^2-4)*1/2*1/(x^2-4)^(1/2)*2x
y'=1/sqrt(x^2-4)*1/2*1/sqrt(x^2-4)*2x
y'=x/(x^2-4)

Therefore, the derivative of the given function is y'=x/(x^2-4) .