Problem 67 - Thomas Calculus

Question

In Exercises \(57-70\) , 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

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

Step 1: Take the Natural Logarithm

Take the natural logarithm of both sides of the equation. This allows us to use the properties of logarithms to simplify the expression. We set \( y = \frac{x \sqrt{x^2+1}}{(x+1)^{2/3}} \) and take \( \ln(y) = \ln\left(\frac{x \sqrt{x^2+1}}{(x+1)^{2/3}}\right) \). Hence, \( \ln(y) = \ln(x) + \ln(\sqrt{x^2+1}) - \ln((x+1)^{2/3}) \).

Step 2: Simplify Using Logarithm Properties

Simplify the expression by using logarithmic identities: \( \ln(x) + \frac{1}{2}\ln(x^2+1) - \frac{2}{3}\ln(x+1) \). The identities used include \( \ln(a) + \ln(b) = \ln(a \cdot b) \) and \( \ln(a^b) = b\ln(a) \).

Step 3: Differentiate with Respect to x

Differentiate both sides of the equation with respect to \( x \). Using implicit differentiation on the left, we have \( \frac{1}{y} \frac{dy}{dx} \). The right side gives: \( \frac{d}{dx}\left(\ln(x)\right) + \frac{1}{2} \frac{d}{dx}\left(\ln(x^2+1)\right) - \frac{2}{3} \frac{d}{dx}\left(\ln(x+1)\right) \).

Step 4: Compute the Derivatives

Now compute each derivative separately: \( \frac{d}{dx}(\ln(x)) = \frac{1}{x} \), \( \frac{1}{2} \frac{d}{dx}(\ln(x^2+1)) = \frac{x}{x^2+1} \), and \(-\frac{2}{3} \frac{d}{dx}(\ln(x+1)) = -\frac{2}{3(x+1)} \).

Step 5: Substitute and Solve for \(\frac{dy}{dx}\)

Substituting back into the differentiated equation: \( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \frac{x}{2(x^2+1)} - \frac{2}{3(x+1)} \). Solve for \( \frac{dy}{dx} \) to get: \( \frac{dy}{dx} = y\left(\frac{1}{x} + \frac{x}{2(x^2+1)} - \frac{2}{3(x+1)}\right) \).

Step 6: Substitute Back the Expression for y

Replace \( y \) with its original expression \( \frac{x \sqrt{x^2+1}}{(x+1)^{2/3}} \) to get \( \frac{dy}{dx} = \frac{x \sqrt{x^2+1}}{(x+1)^{2/3}}\left(\frac{1}{x} + \frac{x}{2(x^2+1)} - \frac{2}{3(x+1)}\right) \).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. It is a fundamental concept in calculus because it helps simplify complex expressions, especially those involving exponential growth or decay.

In logarithmic differentiation, we often use the natural logarithm due to its convenient properties that make differentiation straightforward. These properties are:

• \( \ln(a \cdot b) = \ln(a) + \ln(b) \)
• \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
• \( \ln(a^b) = b \cdot \ln(a) \)

By taking the natural logarithm of both sides of an equation, we can transform multiplication and division into addition and subtraction, simplifying differentiation tasks.

Implicit Differentiation

Implicit differentiation is a powerful tool used in calculus when dealing with equations that define functions implicitly rather than explicitly.

In explicit differentiation, the function \( y \) is directly written in terms of \( x \), making it easy to apply rules like the power rule or product rule. However, implicit differentiation comes in handy when we can't easily solve for \( y \).

When we take the derivative, we differentiate both sides of an equation, applying differentiation rules while treating \( y \) as a function of \( x \). We then use the chain rule, taking \( \frac{dy}{dx} \) into account whenever we differentiate a term that contains \( y \). This way, we find the relation between \( \frac{dy}{dx} \) and \( x \) despite not having \( y \) explicitly expressed as a function of \( x \).

In our specific example, implicit differentiation is utilized once the natural logarithm has been applied, making it easier to differentiate the resulting equation.

Derivative of Functions

Derivatives measure the rate at which a quantity changes. In calculus, it is fundamental to find the derivative of functions to understand how they behave or to find maxima and minima.

When differentiating functions, there are several essential rules to consider:

• The power rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \)
• Derivative of \( \ln(x) \) is \( \frac{1}{x} \)
• Chain rule: If a function \( g \) is applied to \( x \), then its derivative is \( g'(f(x)) \cdot f'(x) \)

These rules are applied in step-by-step procedures to compute derivatives, especially when dealing with complex expressions simplified through logarithmic differentiation.

In our exercise, the derivatives of logarithmic expressions \( \ln(x) \), \( \ln(x^2+1) \), and \( \ln(x+1) \) are individually calculated and combined to provide the final result.

Logarithmic Identities

Logarithmic identities are mathematical tools that aid in simplifying complex equations, especially when tackling problems involving exponents.

The main identities are:

• \( \ln(a \cdot b) = \ln(a) + \ln(b) \)
• \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
• \( \ln(a^b) = b \cdot \ln(a) \)

These identities transform products into sums, quotients into differences, and exponents into multipliers, thereby streamlining differentiation processes.

By applying these identities before differentiation, the expression can be simplified, making it easier to find the derivative. For instance, in our problem, the initial application of these identities helps in rewriting the function into simpler terms before using implicit differentiation. This highlights the valuable role of logarithmic identities in calculus.