Question

For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$ y=\sqrt{x^{2}+1} \sqrt{x^{2}-1} $$

Short answer

\( \frac{dy}{dx} = \sqrt{x^2 + 1} \sqrt{x^2 - 1} \left(\frac{x}{x^2 + 1} + \frac{x}{x^2 - 1}\right) \).

Step-by-step solution

Express in Logarithmic Form

First, express the given function using natural logarithms for easier differentiation. We have:\[ y = \sqrt{x^2 + 1} \sqrt{x^2 - 1} \]Taking the natural logarithm of both sides:\[ \ln y = \ln(\sqrt{x^2 + 1}) + \ln(\sqrt{x^2 - 1}) \].

Simplify the Logarithmic Expression

Use the logarithmic identity \( \ln\sqrt{a} = \frac{1}{2} \ln a \) to simplify:\[\ln y = \frac{1}{2} \ln(x^2 + 1) + \frac{1}{2} \ln(x^2 - 1)\].This separates the logarithm based on the properties of logarithms.

Differentiate Implicitly

Differentiate both sides with respect to \(x\) using implicit differentiation.For the left side: \( \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} \).For the right side, apply the chain rule to each term:\[\frac{1}{2}\frac{d}{dx}\ln(x^2 + 1) + \frac{1}{2}\frac{d}{dx}\ln(x^2 - 1) = \frac{1}{2} \cdot \frac{2x}{x^2 + 1} + \frac{1}{2} \cdot \frac{2x}{x^2 - 1}\].

Solve for \( \frac{dy}{dx} \)

Simplify the expression:\[\frac{1}{y}\frac{dy}{dx} = \frac{x}{x^2 + 1} + \frac{x}{x^2 - 1}\].Multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):\[\frac{dy}{dx} = y\left(\frac{x}{x^2 + 1} + \frac{x}{x^2 - 1}\right)\].Substitute \(y = \sqrt{x^2 + 1} \sqrt{x^2 - 1}\) back into the equation:\[\frac{dy}{dx} = \sqrt{x^2 + 1} \sqrt{x^2 - 1}\left(\frac{x}{x^2 + 1} + \frac{x}{x^2 - 1}\right)\].

Key concepts

Logarithmic Differentiation

Logarithmic differentiation is a clever method used when differentiating complex functions, particularly those involving products, quotients, or powers. When a function is difficult to differentiate in its given form, taking the natural logarithm of both sides can transform the function into a more manageable form. This technique is very useful when one can apply the properties of logarithms to simplify the expression.

The process involves taking the natural logarithm, applying the properties of logarithms (such as \(\ln(ab) = \ln a + \ln b\) and \(\ln(a^b) = b \ln a\)), simplifying the expression, and finally differentiating both sides. This approach converts multiplication into addition and divisions into subtraction, making differentiation easier. In our exercise, logging the original function allowed us to break down the product of square roots into a sum, which is crucial when dealing with products of functions.

• Take the natural logarithm of both sides.
• Apply logarithmic identities to simplify the expression.
• Differentiate the simplified expression.

Chain Rule

The chain rule is a fundamental differentiation rule used to differentiate composite functions, which are functions within functions. When a variable is nested inside another function, the chain rule becomes essential.

Using the chain rule, one must differentiate the outer function (leaving the inside unchanged), then multiply by the derivative of the inner function. This rule surfaces in our step-by-step solution when dealing with expressions like \(\ln(x^2 + 1)\), where \(x^2 + 1\) is the inner function of the natural log. The derivative of a natural log function is \(\frac{1}{u}\) times the derivative of \(u\), where \(u\) is the inner function.

• Identify the inner and outer functions.
• Ddifferentiate the outer function with the inner function still inside.
• Multiply by the derivative of the inner function.

Product Rule

The product rule is essential when taking the derivative of expressions involving the product of two functions. It ensures that each function and its respective derivative are correctly accounted for, giving an accurate result when differentiating.

In the context of our exercise, the original function was the product of two square roots. By converting the product into a sum of logarithms, we simplify a potentially complex product differentiation into a series of easier individual differentiations. In typical scenarios, the product rule states that if you have two functions \(u(x)\) and \(v(x)\), the derivative of their product is given as \(u'v + uv'\). However, applying logarithmic differentiation here makes it more straightforward by using the properties of logarithms to convert the multiplication context into one of addition.

• Logarithmic differentiation bypasses the traditional product rule.
• Simplifies the differentiation of product functions.

Natural Logarithm

The natural logarithm is a special logarithm with a base of \(e\), where \(e\) is approximately equal to 2.71828. It's particularly useful in calculus because it possesses unique properties that simplify the process of differentiation and integration.

In mathematical notation, the natural logarithm of a number \(x\) is denoted as \(\ln x\). One of its key properties, \(\frac{d}{dx}(\ln x) = \frac{1}{x}\), aids in making differentiation involving logarithms straightforward. This property is frequently used in combination with the chain rule when differentiating composite functions. In our exercise, the natural logarithm allowed us to transform complex multiplicative expressions into simpler additive ones, facilitating the differentiation process.

• Useful in simplifying differentiation.
• Has a derivative of \(\frac{1}{x}\).
• Facilitates the breakdown of complex expressions.