Problem 111 - Calculus and its applications

Question

Differentiate. $$ f(t)=\ln \left|\frac{1-t}{1+t}\right| $$

Step-by-Step Solution

Answer

The derivative is \( f'(t) = \frac{-2}{1-t^2} \).

Step 1: Understand the Problem

We need to differentiate the function \( f(t) = \ln \left| \frac{1-t}{1+t} \right| \). This involves using the chain rule and the derivative of logarithmic functions.

Step 2: Rewrite the Function

Notice that \( \ln \left| \frac{1-t}{1+t} \right| \) can be expressed as \( \ln \left( \frac{1-t}{1+t} \right) \) since we are differentiating with respect to \( t \), assuming \( 1-t eq 0 \) and \( 1+t eq 0 \).

Step 3: Use the Chain Rule

Let \( g(t) = \frac{1-t}{1+t} \). Then \( f(t) = \ln |g(t)| \), and we apply the chain rule to find \( f'(t) = \frac{1}{g(t)} \cdot g'(t) \).

Step 4: Differentiate \( g(t) \)

We need \( g'(t) \). Use the quotient rule: \( g'(t) = \frac{(1+t)(-1) - (1-t)(1)}{(1+t)^2} = \frac{-1-t - (1-t)}{(1+t)^2} = \frac{-2}{(1+t)^2} \).

Step 5: Combine Results

Substitute \( g(t) \) and \( g'(t) \) in the expression from Step 3: \( f'(t) = \frac{1}{\frac{1-t}{1+t}} \cdot \frac{-2}{(1+t)^2} = \frac{-2(1+t)}{(1-t)(1+t)^2} = \frac{-2}{(1-t)(1+t)} \).

Step 6: Simplify the Expression

Since the expression simplifies to \( \frac{-2}{1-t^2} \), we have found the derivative of the function given. Note \( f'(t) = \frac{-2}{1-t^2} \).

Key Concepts

Chain Rule

The Chain Rule is an essential technique in calculus used for differentiating composite functions. In simple terms, when you have a function composed of two other functions, you apply the Chain Rule to differentiate.
For example, if you have a function like \( f(x) = h(g(x)) \), the derivative \( f'(x) \) is found by multiplying the derivative of the outer function \( h \) evaluated at the inner function \( g \), with the derivative of the inner function \( g \). This gives us \( f'(x) = h'(g(x)) \cdot g'(x) \).
In the given problem, we apply the Chain Rule when finding the derivative of \( f(t) = \ln |g(t)| \), where \( g(t) = \frac{1-t}{1+t} \).

• The derivative of \( \ln |x| \) is \( \frac{1}{x} \).
• Using the chain rule, the derivative \( f'(t) = \frac{1}{g(t)} \cdot g'(t) \).

This rule simplifies the process and ensures precision in evaluating derivatives of complex, composite functions.

Quotient Rule

The Quotient Rule is utilized when differentiating functions that are ratios of two expressions. Knowing how to apply this rule is crucial whenever you face a division situation in functions.
For a function expressed as a quotient, such as \( g(x) = \frac{u(x)}{v(x)} \), the derivative is:
\[ g'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]
Applying this rule ensures you accurately account for changes in both the numerator and the denominator functions.
In our example, \( g(t) = \frac{1-t}{1+t} \). Following the Quotient Rule:

• \( u(t) = 1-t \) and \( u'(t) = -1 \).
• \( v(t) = 1+t \) and \( v'(t) = 1 \).
• Thus, \( g'(t) = \frac{(1+t)(-1) - (1-t)(1)}{(1+t)^2} = \frac{-2}{(1+t)^2} \).

This provides an efficient way to handle differentiations involving fractions.

Logarithmic Differentiation

Logarithmic Differentiation is a powerful technique, especially useful for differentiating functions with complex structures or products and quotients of exponential expressions. By leveraging the properties of logarithms, complex multiplication and division are transformed into addition and subtraction, thereby easing the differentiation process.
Consider a function like \( y = \ln \left| \frac{1-t}{1+t} \right| \) from our problem. Using logarithmic differentiation:

• First, simplify by rewriting as \( y = \ln \left( \frac{1-t}{1+t} \right) \), assuming that \( 1-t eq 0 \) and \( 1+t eq 0 \).
• Next, differentiate the expression using the Chain Rule: \( y' = \frac{1}{g(t)} \cdot g'(t) \).

This method capitalizes on the simpler properties of logarithms to differentiate, particularly useful in our exercise where the natural log function is involved. The result \( f'(t) = \frac{-2}{1-t^2} \) elegantly captures the derivative of the original function using this technique.