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The chain rule is a fundamental concept in calculus used to find the derivative of composite functions. When dealing with a function like \( f(x) = \sqrt{x^2-1} \), you are essentially looking at a composition of functions. The outer function is the square root, \( \sqrt{u} \), and the inner function is \( u = x^2 - 1 \). To find the derivative using the chain rule, you take the derivative of the outer function and multiply it by the derivative of the inner function. Here's how it breaks down step by step:

• The derivative of \( \sqrt{u} \) is \( \frac{1}{2\sqrt{u}} \).
• The derivative of the inner function \( u = x^2 - 1 \) is \( 2x \).

Thus, applying the chain rule, we get the derivative \( f'(x) = \frac{1}{2\sqrt{x^2-1}} \times 2x = \frac{x}{\sqrt{x^2-1}} \). This step is crucial to understand how the derivative \( f'(x) \) gives us insight into the behavior of the function, such as its critical points and intervals of increase and decrease.

Critical points are values of \( x \) where the derivative of a function is either zero or undefined, indicating potential local maxima, minima, or points of inflection. When finding the critical points for \( f(x) = \sqrt{x^2-1} \), we examine \( f'(x) = \frac{x}{\sqrt{x^2-1}} \).First, consider where the derivative is zero. This happens when the numerator, \( x \), is zero. However, substituting \( x = 0 \) into the original function \( f(x) \) yields a non-real number, indicating no critical points here due to undefined function. Next, check where the derivative is undefined. That happens when the denominator is zero, i.e., \( \sqrt{x^2-1} = 0 \), simplifying to \( x^2 = 1 \) or \( x = \pm 1 \). Yet both \( x = 1 \) and \( x = -1 \) fall on the boundary or outside the domain of \( f(x) \), thereby producing no valid critical points in the interval where \( f(x) \) exists.

Analyzing intervals of increase and decrease involves looking at where the derivative of the function is positive (increasing) or negative (decreasing). For the function \( f(x) = \sqrt{x^2-1} \), its derivative \( f'(x) = \frac{x}{\sqrt{x^2-1}} \) gives us valuable insight. Given that there are no critical points, the next step is to examine the sign of \( f'(x) \) over its domain. Consider examining regions divided by critical points and other key markers:

• For \( x > 1 \), \( f'(x) > 0 \). This tells us the function is increasing in this interval.
• For \( x < -1 \), \( f'(x) < 0 \). Hence, the function is decreasing, but only relevant if the domain allows consideration here; otherwise modern consideration stays in valid domains.

Ultimately, due to the nature of \( f(x) \) having domain constraints, the only practical interval where \( f(x) \) increases is when \( x > 1 \). Thus, the conclusion is that \( f(x) \) is increasing on \( (1, \infty) \), giving a full picture of behavioral tendencies over this range.