Problem 58 - Thomas Calculus

Question

In Exercises \(53-58,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\) . $$ f(u)=\left(\frac{u-1}{u+1}\right)^{2}, \quad u=g(x)=\frac{1}{x^{2}}-1, \quad x=-1 $$

Step-by-Step Solution

Answer

The value of \((f \circ g)'(-1)\) is \(-8\).

Step 1: Identify Functions and Derivatives

First, identify the functions involved and their derivatives. We have \( f(u) = \left( \frac{u-1}{u+1} \right)^2 \) and \( u = g(x) = \frac{1}{x^2} - 1 \). The goal is to find \((f \circ g)'(x)\). To do this, we'll need to use the chain rule \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x)\).

Step 2: Differentiate \(f(u)\)

To find \( f'(u) \), use the chain rule. Let \( h(u) = \frac{u-1}{u+1} \). Then, \( f(u) = h(u)^2 \), so \( f'(u) = 2h(u) \cdot h'(u) \). Find \( h'(u) \) using the quotient rule: \( h'(u) = \frac{(u+1) \cdot 1 - (u-1) \cdot 1}{(u+1)^2} = \frac{2}{(u+1)^2} \). Thus, \( f'(u) = 2 \cdot \frac{u-1}{u+1} \cdot \frac{2}{(u+1)^2} = \frac{4(u-1)}{(u+1)^3} \).

Step 3: Differentiate \(g(x)\)

Differentiate \( g(x) = \frac{1}{x^2} - 1 \). The derivative is \( g'(x) = -\frac{2}{x^3} \), as the derivative of \( \frac{1}{x^2} \) is \( -\frac{2}{x^3} \), and the derivative of \(-1\) is zero.

Step 4: Compute \( (f \circ g)'(x) \)

Now substitute \( g(x) \) into \( f'(u) \). At \( x = -1 \), we have \( u = g(-1) = \frac{1}{(-1)^2} - 1 = 0 \). Therefore, \( f'(g(-1)) = \frac{4(0-1)}{(0+1)^3} = -4 \). Also, \( g'(-1) = -\frac{2}{(-1)^3} = 2 \). Thus, \( (f \circ g)'(-1) = f'(g(-1)) \cdot g'(-1) = -4 \cdot 2 = -8 \).

Step 5: Conclusion

The value of \( (f \circ g)'(x) \) at \( x = -1 \) is \(-8\).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental tool in calculus used to describe how a function changes at any point. They represent the slope of the tangent line to the graph of the function. This concept is significant because it helps us understand rates of change and allows us to predict the behavior of functions over small intervals.

When you're working with a function, finding its derivative involves using certain rules and formulas, such as:

• The power rule, where if you have a term like \( x^n \), its derivative is \( n \cdot x^{n-1} \).
• Product and quotient rules, which help when dealing with functions that are the multiplication or division of two simpler functions.

In the particular problem we're examining, the derivatives of the functions \( f(u) \) and \( g(x) \) were essential in calculating the overall derivative of the composite function. The step-by-step solution involved using the quotient rule to find the derivative of a rational function and the power rule to manage polynomial expressions.

Concept of Function Composition

Function composition, denoted as \( (f \circ g)(x) = f(g(x)) \), is a process of applying one function to the results of another. It’s like putting one function inside another, creating a function within a function, which can seem a bit complex at first glance.

In this exercise, function composition meant starting with an initial function \( g(x) \) and then applying the function \( f(u) \) to the result of \( g(x) \). The key is to transform \( x \) through \( g(x) \), and then use that output, \( u \), as the input for \( f(u) \).

Understanding composition is crucial when it comes to calculus because many complex functions can be broken down into simpler, more manageable pieces, allowing us to use calculus techniques effectively on each component. This makes solving derivatives and integrals more straightforward when dealing with composite functions.

Steps in Calculus Problem Solving

Solving calculus problems often involves a clear, logical sequence of steps to ensure accuracy and efficiency. For composite functions, this includes:

• Identifying and understanding each function involved and any given values.
• Calculating the derivative of each individual function, applying rules like the power rule, chain rule, or quotient rule as necessary.
• Substituting the given values into the derivatives to find specific values at particular points.

In the exercise provided, these steps were followed meticulously. First, the derivatives of \( f(u) \) and \( g(x) \) were calculated using appropriate rules. Next, the chain rule was used for the derivative of the composite function. Substituting the value of \( x \), determined from \( g(-1) \), was crucial for calculating the final derivation value.

Approaching calculus problems methodically by breaking them into smaller tasks makes complex problems more manageable and solvable.