Problem 65 - Thomas Calculus in SI Units

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

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

Step 1: Apply Natural Logarithm

Take the natural logarithm of both sides of the equation to obtain: \( \ln(y) = \ln\left(\frac{x \sqrt{x^2+1}}{(x+1)^{2/3}}\right) \).

Step 2: Use Logarithm Properties

Use the properties of logarithms to simplify the expression:\( \ln(y) = \ln(x) + \ln(\sqrt{x^2+1}) - \ln((x+1)^{2/3}) \).

Step 3: Simplify With Additional Log Properties

Further, simplify the terms: \( \ln(y) = \ln(x) + \frac{1}{2} \ln(x^2+1) - \frac{2}{3} \ln(x+1) \).

Step 4: Differentiate Implicitly

Differentiate both sides with respect to \(x\):\( \frac{d}{dx}(\ln(y)) = \frac{d}{dx}(\ln(x) + \frac{1}{2}\ln(x^2+1) - \frac{2}{3}\ln(x+1)) \).Use the chain rule: \( \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} + \frac{1}{2} \cdot \frac{2x}{x^2+1} - \frac{2}{3} \cdot \frac{1}{x+1} \).

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

Rearrange to solve for \(\frac{dy}{dx}\):\( \frac{dy}{dx} = y\left( \frac{1}{x} + \frac{x}{x^2+1} - \frac{2}{3(x+1)} \right) \).

Step 6: Substitute \(y\) Back

Substitute \(y = \frac{x \sqrt{x^2+1}}{(x+1)^{2/3}} \) into the derivative expression:\( \frac{dy}{dx} = \frac{x \sqrt{x^2+1}}{(x+1)^{2/3}}\left( \frac{1}{x} + \frac{x}{x^2+1} - \frac{2}{3(x+1)} \right) \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. In essence, it allows us to differentiate a function that is nested within another function. The Chain Rule states that if you have a function \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is given by \( f'(g(x)) \cdot g'(x) \).
To apply this rule, we take the derivative of the outer function evaluated at the inner function, and multiply it by the derivative of the inner function.
For instance, when differentiating \( \ln(f(x)) \), the Chain Rule helps us differentiate the inside function, \( f(x) \), as well.- When we dealt with the term \( \frac{1}{2} \ln(x^2+1) \) in the original exercise, we first differentiate \( \ln(u) \) to get \( \frac{1}{u} \) and then use the Chain Rule to multiply by \( \frac{d}{dx}(x^2+1) = 2x \).- This resulted in \( \frac{1}{2} \cdot \frac{2x}{x^2+1} \).
This process ensures each "layer" of the function is considered. Remember, derivatives involve breaking down functions layer by layer, and that's precisely what the Chain Rule facilitates.

Implicit Differentiation

Implicit differentiation is a valuable technique used when functions are not given in the form \( y = f(x) \) and instead are intertwined together. In cases where solving for \( y \) explicitly is difficult or impossible, implicit differentiation becomes invaluable.
In general, you treat \( y \) as an implicit function of \( x \) and differentiate both sides of the equation with respect to \( x \), keeping in mind that whenever you differentiate \( y \), you'll need to multiply by \( \frac{dy}{dx} \). This arises quite often in logarithmic differentiation because taking logs often turns explicit functions into implicit forms.
In the given exercise, \( \ln(y) \) represents differentiating implicitly since \( y \) is not isolated. After applying the logarithmic properties, we differentiate \( \ln(y) \), leading to the derivative \( \frac{1}{y} \frac{dy}{dx} \). Solving involves manipulating and isolating \( \frac{dy}{dx} \) for the final derivative.

Properties of Logarithms

Logarithmic properties simplify expressions and make differentiation easier. Understanding these properties allows complex expressions to be broken down into more manageable parts. There are three main properties frequently used:

• Product Rule for Logarithms: \( \ln(a \cdot b) = \ln a + \ln b \).
• Quotient Rule for Logarithms: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \).
• Power Rule for Logarithms: \( \ln(a^b) = b \ln a \).

These rules were applied in the exercise to break down the expression and make differentiation straightforward.
For example, we transformed \( \ln\left(\frac{x \sqrt{x^2+1}}{(x+1)^{2/3}}\right) \) using these properties: - First, recognize the quotient and apply the Quotient Rule: convert it into \( \ln(x) + \ln(\sqrt{x^2+1}) - \ln((x+1)^{2/3}) \).- Then, simplify further using the Power Rule for logs: \( \ln(x) + \frac{1}{2}\ln(x^2+1) - \frac{2}{3}\ln(x+1) \).By doing so, you ease the differentiation process since the properties of logarithms reduce complexity and turn multiplication into addition, division into subtraction, and powers into coefficients, making implicit differentiation more accessible.