Problem 44 - Calculus

Question

Verify the derivatives , which give useful anti derivatives: $$ \frac{d}{d x} \ln \left(\frac{x-a}{x+a}\right)=\frac{2 a}{\left(x^{2}-a^{2}\right)} $$

Step-by-Step Solution

Answer

The derivative is verified as \( \frac{2a}{x^2-a^2} \).

Step 1: Differentiate the given function

First, we'll take the function inside the natural logarithm, \( \ln \left(\frac{x-a}{x+a}\right) \), and differentiate it with respect to \(x\). According to the chain rule, the derivative of \( \ln(u) \) with respect to \(x\) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \frac{x-a}{x+a} \).

Step 2: Differentiate the inner function using the quotient rule

To find \( \frac{du}{dx} \), use the quotient rule for differentiating \( \frac{x-a}{x+a} \). The quotient rule states \( \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \). Set \( f(x) = x-a \) and \( g(x) = x+a \).

Step 3: Compute the derivatives of numerator and denominator

Calculate \( f'(x) = \frac{d}{dx}(x-a) = 1 \) and \( g'(x) = \frac{d}{dx}(x+a) = 1 \). Using the quotient rule, calculate: \(\frac{du}{dx} = \frac{(1)(x+a) - (x-a)(1)}{(x+a)^2} = \frac{x+a - x + a}{(x+a)^2} = \frac{2a}{(x+a)^2}.\)

Step 4: Substitute into derivative of logarithm function

Substitute \( u = \frac{x-a}{x+a} \) and \( \frac{du}{dx} = \frac{2a}{(x+a)^2} \) into the derivative formula \( \frac{1}{u} \cdot \frac{du}{dx} \). Substituting gives us: \[\frac{d}{dx} \ln \left(\frac{x-a}{x+a}\right) = \frac{1}{\frac{x-a}{x+a}} \cdot \frac{2a}{(x+a)^2} = \frac{(x+a) \cdot 2a}{(x-a)(x+a)^2}.\]

Step 5: Simplify the expression

Simplify the expression: \( \frac{(x+a) \cdot 2a}{(x-a)(x+a)^2} = \frac{2a(x+a)}{(x^2-a^2)(x+a)} \). Cancel \( x+a \) in the numerator and denominator to achieve: \( \frac{2a}{x^2-a^2} \), which confirms the given derivative.

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. Imagine a composite function as a function nested within another function, like an onion with layers. You start by peeling one layer to understand its derivative. The chain rule helps with this by breaking down the process step by step.

The chain rule states that if you have a function, say \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is found by multiplying the derivative of the outer function evaluated at the inner function \( f'(g(x)) \), by the derivative of the inner function \( g'(x) \). So, the formula looks like:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \)

In the exercise, we apply the chain rule to differentiate the natural logarithm of a fraction. We set \( u = \frac{x-a}{x+a} \) and then apply the chain rule with respect to \( u \). This helps us manage the complexity by focusing on one function at a time.

Quotient Rule

When working with functions that are the quotient, or division, of two simpler functions, the quotient rule becomes your ally. It's a specific method to find the derivative of a function that can be written as \( \frac{f(x)}{g(x)} \), where both \( f(x) \) and \( g(x) \) are differentiable.

To remember, there's a handy rhyme: *"Low 'D' high, minus high 'D' low, square the bottom, away we go"*. Mathematically, this relationship is expressed as:

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

In our exercise, this helped differentiate \( \frac{x-a}{x+a} \). We took the derivatives of the numerator \( (x-a) \) and denominator \( (x+a) \) separately. Then we applied the rule which resulted in \( \frac{2a}{(x+a)^2} \). This derivative becomes crucial when substituted back into the chain rule application.

Natural Logarithm Differentiation

Differentiating natural logarithm functions is straightforward due to their unique properties. The function \( \ln(x) \) has a derivative of \( \frac{1}{x} \). This baseline rule extends neatly to composite functions, where adjustments are made using the chain rule.

When dealing with a natural logarithm of a complex function like \( \ln(u) \), where \( u \) itself is a function of \( x \), the differentiation becomes \( \frac{1}{u} \cdot \frac{du}{dx} \).

In our provided exercise, the function inside the logarithm, generated by a fraction, required additional steps of differentiation. Once we obtained \( \frac{du}{dx} \) through the quotient rule, substituting it gave us a derivative consistent with the natural properties of logarithms. Hence, the final simplification relied on understanding each component’s derivative behavior.