Problem 34 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\frac{1}{2} \ln \frac{1+x}{1-x}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{1-x^2} \).

Step 1: Identify the Function and Rule Needed

The given function is \( y = \frac{1}{2} \ln \frac{1+x}{1-x} \). To find the derivative of this function with respect to \(x\), we will use the chain rule and the derivative of the natural logarithm rule.

Step 2: Differentiate the Natural Logarithm

The derivative of \( \ln u \) with respect to \(x\) is \( \frac{1}{u} \frac{du}{dx} \). For \( u = \frac{1+x}{1-x} \), we need to find \( \frac{du}{dx} \).

Step 3: Differentiate the Inner Function \(u\)

Since \( u = \frac{1+x}{1-x} \), apply the quotient rule: \( \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2} \). Here, \( f = 1+x \) and \( g = 1-x \).

Step 4: Apply the Quotient Rule

Compute \( f' = 1 \) and \( g' = -1 \). Then:\[ \frac{du}{dx} = \frac{(1)(1-x) - (1+x)(-1)}{(1-x)^2} \]Simplify:\[ \frac{du}{dx} = \frac{1-x + 1 + x}{(1-x)^2} = \frac{2}{(1-x)^2} \]

Step 5: Substitute and Simplify

Substitute \( \frac{du}{dx} \) back into:\[ \frac{d}{dx}\left( \ln \frac{1+x}{1-x} \right) = \frac{1}{\frac{1+x}{1-x}} \cdot \frac{2}{(1-x)^2} \]This simplifies to:\[ \frac{2(1-x)}{(1+x)(1-x)} = \frac{2}{1-x^2} \]

Step 6: Multiply by Constant

The given function \( y = \frac{1}{2} \ln \frac{1+x}{1-x} \) includes the constant \( \frac{1}{2} \), so multiply the derivative by it:\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{2}{1-x^2} = \frac{1}{1-x^2} \]

Key Concepts

Chain Rule

The chain rule is a fundamental theorem in calculus used to compute the derivative of a composite function. It states that if we have a function composed of two functions, say, \(y = f(g(x))\), the derivative of \(y\) with respect to \(x\) is given by multiplying the derivative of \(f\) with respect to \(g(x)\) by the derivative of \(g(x)\) with respect to \(x\). In formula terms:

• \(\frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\)

For the exercise at hand, the chain rule is used to differentiate \(y = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\). Here, the logarithmic function is applied to a more complex expression, \(\frac{1+x}{1-x}\), our 'inner function'.
By applying the chain rule, we first find how the natural log function depends on its inner function and then how this inner function depends on \(x\). This understanding is crucial as it allows us to break down complex derivatives into more manageable steps.

Logarithmic Differentiation

Logarithmic differentiation is a method involving the natural logarithm to simplify differentiation of complicated functions. When faced with a function like \(y = \ln u\), where \(u\) itself is a function of another variable \(x\), the derivative \(\frac{d}{dx}(\ln u)\) depends on applying the chain rule after initially rewriting as \(\frac{1}{u} \cdot \frac{du}{dx}\).

• The natural logarithm simplifies multiplication into addition, division into subtraction, and exponentiation into multiplication, making derivatives less cumbersome.

In our example, the differentiation of \(\ln\left(\frac{1+x}{1-x}\right)\) showcases how logarithmic differentiation makes dealing with the division in the argument of the logarithm more straightforward.
First, compute \(\frac{du}{dx}\) and then simplify as demonstrated: \(\frac{d}{dx}\left(\ln\left(\frac{1+x}{1-x}\right)\right) = \frac{1}{\frac{1+x}{1-x}} \cdot \frac{du}{dx}\). This step sets up the simplified function for final evaluation by plugging in \(\frac{du}{dx}\).

Quotient Rule

The quotient rule is crucial when differentiating functions expressed as one function divided by another. For a function \(u = \frac{f(x)}{g(x)}\), the derivative \(u'\) is computed using:

• \(\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2}\)

In the original exercise, \(u\) is represented by \(\frac{1+x}{1-x}\). We apply the quotient rule to find \(\frac{du}{dx}\), where:

• \(f(x) = 1+x\) with derivative \(f'(x) = 1\)
• \(g(x) = 1-x\) with derivative \(g'(x) = -1\)

Substituting into the quotient rule formula, we get:

• \(\frac{du}{dx} = \frac{(1)(1-x) - (1+x)(-1)}{(1-x)^2}\)

Simplifying this gives \(\frac{du}{dx} = \frac{2}{(1-x)^2}\), an important step that feeds into the differentiation process for our composite function. This process illustrates the powerful utility of the quotient rule in calculus, efficiently handling division in derivatives.