Problem 51 - University Calculus: Early Transcendentals

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\frac{x \sqrt{x^{2}+1}}{(x+1)^{2 / 3}}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \frac{x \sqrt{x^2 + 1}}{(x+1)^{2/3}} \left( \frac{1}{x} + \frac{x}{x^2+1} - \frac{2}{3(x+1)} \right) \).

Step 1: Logarithmic Transformation

Begin by taking the natural logarithm of both sides of the equation to simplify the differentiation process. Start with the equation: \[ y = \frac{x \sqrt{x^{2}+1}}{(x+1)^{2/3}} \] After taking the natural logarithm, it becomes: \[ \ln(y) = \ln\left(\frac{x \sqrt{x^2 + 1}}{(x + 1)^{2/3}}\right) \] Use logarithmic properties to split this into: \[ \ln(y) = \ln(x) + \ln(\sqrt{x^2 + 1}) - \ln((x + 1)^{2/3}) \] Further simplify to: \[ \ln(y) = \ln(x) + \frac{1}{2}\ln(x^2 + 1) - \frac{2}{3}\ln(x + 1) \]

Step 2: Differentiate with Respect to x

Differentiate both sides of the equation with respect to \(x\). On the left side, use the chain rule: \[ \frac{d}{dx}[\ln(y)] = \frac{1}{y} \cdot \frac{dy}{dx} \] For the right side: - Differentiate \(\ln(x)\) to get \(\frac{1}{x}\). - For \(\frac{1}{2}\ln(x^2 + 1)\), use the chain rule: \[ \frac{1}{2} \cdot \frac{1}{x^2 + 1} \cdot 2x = \frac{x}{x^2+1} \] - For \(-\frac{2}{3}\ln(x+1)\), use the chain rule: \[ -\frac{2}{3} \cdot \frac{1}{x+1} = -\frac{2}{3(x+1)} \] Combine these to find: \[ \frac{dy}{dx} = y \left( \frac{1}{x} + \frac{x}{x^2+1} - \frac{2}{3(x+1)} \right) \]

Step 3: Substitute Back for y

Replace \(y\) with its original expression in the derivative formula found: \[ \frac{dy}{dx} = \frac{x \sqrt{x^2 + 1}}{(x+1)^{2/3}} \left( \frac{1}{x} + \frac{x}{x^2+1} - \frac{2}{3(x+1)} \right) \] Conduct basic simplifications to ensure each term is as concise as possible, as needed.

Key Concepts

Derivative

A derivative represents how a function changes as its input changes. It shows the rate at which a quantity, like distance or temperature, transforms over time or space.
In calculus, when we speak about finding the derivative of a function, we're essentially calculating this rate of change. To do this, we follow a defined process:

• Identify the function to be differentiated.
• Apply differentiation rules, like the power, product, or quotient rules, to find its derivative.
• Simplify the result to its simplest form.

In the context of the exercise, we used logarithmic differentiation as a method to simplify the process. This technique is helpful when dealing with complicated functions, like the one given in the exercise, where direct differentiation might be cumbersome.

Chain Rule

The chain rule is a fundamental technique in calculus used to compute the derivative of composite functions. Imagine you have a function nested inside another function. The chain rule helps you "unravel" this nest to find how the overall function changes with respect to its inner functions.For any two functions, if you have a function written as a composition, such as outer function (\(h(x)\)) of an inner function (\(g(x)\)), the chain rule states:\[\frac{d}{dx} h(g(x)) = h'(g(x)) \cdot g'(x)\]This means you take the derivative of the outer function evaluated at the inner function and multiply by the derivative of the inner function.In our exercise, the chain rule helps simplify terms like \(\ln(x^2+1)\). This term involves a composite function where the chain rule allows us to differentiate easily by handling the inside \(x^2 + 1\) first, and then working outwards to the logarithmic part.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a logarithm to the base \(e\), where \(e\) is an irrational number approximately equal to 2.71828. This function is crucial in various areas of math and science, often because of how naturally it occurs in growth processes, such as population growth or radioactive decay.When you apply the natural logarithm to a problem, such as in logarithmic differentiation, it helps simplify products and quotients into sums and differences, making derivative calculation easier.

• Simplifies multiplicative structures to additive ones, via properties such as: \(\ln(ab) = \ln(a) + \ln(b)\).
• Turns powers into products, by using: \(\ln(a^b) = b\ln(a)\).

In our exercise, using the natural logarithm helped break down the complex original function into terms that are easier to differentiate individually.

Differentiation Techniques

Differentiation techniques are strategies or methods employed to find the derivative of functions. Given the diversity of mathematical functions, multiple techniques are available to tackle different situations effectively. For functions that are challenging to differentiate directly, such as the one given, it becomes beneficial to use:

• Logarithmic differentiation: Especially useful when dealing with products or quotients involving exponents or roots, which can be simplified using logarithmic properties.
• Product and Quotient rules: To differentiate functions that are multiplied or divided by one another.
• Chain rule: To handle composite functions effectively, simplifying the differentiation process when a function is composed of another function.

These techniques not only aid in finding derivatives efficiently but also enhance our understanding of how the functions behave, making them indispensable tools in calculus.