Question

In Exercises, find the derivative of the function. $$ y=\ln \sqrt{\frac{x+1}{x-1}} $$

Short answer

The derivative of the function \(y=\ln \sqrt{\frac{x+1}{x-1}}\) is \(y' = \frac{-1}{(x^2 - 1)}\).

Step-by-step solution

Simplify the Function

First, simplify the function to a form easier to manage. Given is \(y=\ln \sqrt{\frac{x+1}{x-1}}\), which equals \(y= \frac{1}{2}\ln ({x+1}) - \frac{1}{2}\ln ({x-1})\). This step is done by breaking up the natural logarithm by employing logarithmic rules.

Differentiating Each Term

After simplifying the function, differentiate each term separately. The derivative of \(y\) will be \(y' = \frac{1}{2}(\frac{1}{{x+1}}) - \frac{1}{2}(\frac{1}{{x-1}})\). Here, the chain rule for derivatives is used (the derivative of ln(x) is 1/x).

Simplify the Result

Finally, simplify the result by finding a common denominator, which is \((x+1)(x-1)\). Therefore, \(y' = \frac{1}{2}\left(\frac{x-1 - (x+1)}{(x+1)(x-1)}\right) = \frac{1}{2}\left(\frac{-2}{(x+1)(x-1)}\right)\). Now it simplifies to \(y' = \frac{-1}{(x^2 - 1)}\).

Key concepts

Natural Logarithm Properties

Understanding the properties of the natural logarithm is essential when differentiating logarithmic functions. The natural logarithm, denoted as \(\ln(x)\), has some distinctive attributes that can simplify complex expressions before taking derivatives.

Arguably, the most powerful of these properties for differentiation purposes are the log of a product, \(\ln(ab) = \ln(a) + \ln(b)\), the log of a quotient, \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\), and the log of a power, \(\ln(a^b) = b\ln(a)\). These properties allow us to break down complicated logarithmic functions into simpler components. For the given exercise \(y=\ln \sqrt{\frac{x+1}{x-1}}\), we utilized the quotient and power rules to transform the expression into \(y=\frac{1}{2}\ln (x+1) - \frac{1}{2}\ln (x-1)\), significantly streamlining the differentiation process.

Chain Rule for Derivatives

When it comes to differentiating composite functions—functions within functions—the chain rule is an indispensable tool. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

In practical terms, if you have a function \(h(x) = f(g(x))\), then its derivative \(h'(x)\) will be \(f'(g(x)) \cdot g'(x)\). Applying this to our problem, where we have to differentiate \(\ln(a(x))\), we consider \(a(x)\) as the inner function. Hence, the derivative \(\frac{d}{dx}\ln(a(x))\) is \(\frac{1}{a(x)} \cdot a'(x)\), where \(a'(x)\) is the derivative of the inner function \(a(x)\). Such understanding streamlines the differentiation of each term in the simplified logarithmic expression.

Simplifying Algebraic Expressions

The goal of simplifying algebraic expressions is to make them more manageable and easier to work with, especially when dealing with complex problems such as derivatives. In the context of our problem, once we've applied the natural logarithm properties and the chain rule for derivatives, we're often left with an expression that, while correct, can be significantly streamlined.

To achieve a cleaner result, we combined like terms and found a common denominator, which, in our case, led to the cancellation of similar terms in the numerator. For the exercise \(y' = \frac{1}{2}(\frac{1}{{x+1}}) - \frac{1}{2}(\frac{1}{{x-1}})\), by finding the common denominator \( (x+1)(x-1)\), we arrive at \(y' = \frac{1}{2}\left(\frac{x-1 - (x+1)}{(x+1)(x-1)}\right)\) which further simplifies to \(y' = \frac{-1}{(x^2 - 1)}\). Effective simplification is not just a matter of aesthetics but also helps in preventing mistakes in further computations and in understanding the behavior of the function.