Problem 80 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(0)=1\) and \(f^{\prime}(0)=2\). Find the derivative of \(f(f(x)-1)\) at \(x=0 .\)

Step-by-Step Solution

Answer

The derivative is 4.

Step 1: Understanding the Chain Rule

To find the derivative of the composite function \( f(f(x)-1) \), we will use the chain rule. The chain rule states that if you have a composite function \( g(h(x)) \), its derivative \( g'(h(x)) \cdot h'(x) \). We will apply this principle to \( f(f(x) - 1) \).

Step 2: Differentiating the Outer Function

Let \( u(x) = f(x) - 1 \). Then the outer function \( f(u) \) has the derivative \( f'(u) \). We need to evaluate this at \( u = f(x) - 1 \).

Step 3: Finding the Inner Function's Derivative

The derivative of \( u(x) = f(x) - 1 \) with respect to \( x \) is simply \( f'(x) \). This is because the derivative of any constant is zero, and so the derivative becomes \( f'(x) - 0 = f'(x) \).

Step 4: Applying the Chain Rule

Using the chain rule, the derivative of \( f(f(x) - 1) \) is \( f'(f(x) - 1) \cdot f'(x) \).

Step 5: Substituting Values to Evaluate at x=0

We evaluate the derivative expression at \( x = 0 \): 1. \( f(x) = f(0) = 1 \) as given.2. So, \( u(0) = f(0) - 1 = 0 \).3. \( f'(x) \), evaluated at \( x=0 \), is \( f'(0) = 2 \).4. So the composite derivative becomes \( f'(u(0)) \times f'(0) = f'(0) \times 2 \).

Step 6: Final Calculation

Substitute the known value of \( f'(0) = 2 \) into the expression to find:\[ f'(0) \times 2 = 2 \times 2 = 4 \]

Key Concepts

Composite Functions Explained

In mathematics, a composite function is a function that is formed when one function is applied and then another function is applied to the result. For example, if we have two functions, \( f(x) \) and \( g(x) \), the composite function \( f(g(x)) \) means that you first apply \( g \) to \( x \) and then apply \( f \) to the result of \( g(x) \).
This is a way to combine multiple functions into a single operation.

Composite functions are represented using the notation \( f(g(x)) \), and the structure involves an outer function and an inner function:

• The **inner function**: In \( f(g(x)) \), \( g(x) \) is the inner function, because it is applied first.
• The **outer function**: \( f(x) \) is the outer function, because it is applied to the result of the inner function.

In our exercise, the function given is \( f(f(x) - 1) \), where \( f(x) - 1 \) acts as the inner function and the second application of \( f \) acts as the outer function. Understanding this composition is key to applying differential rules such as the chain rule.

Derivative Calculation Essentials

When dealing with derivatives, calculating the derivative of a composite function involves using specific rules.
The most common rule for this is the **Chain Rule**. This rule provides a systematic way to differentiate composite functions and is one of the fundamental tools in calculus.

The Chain Rule states:
If you have a composite function \( g(h(x)) \), then its derivative, \( \frac{d}{dx}[g(h(x))] \), is:
\[ g'(h(x)) \cdot h'(x) \]
This formula illustrates that you differentiate the outer function at the inner function and multiply the result by the derivative of the inner function.

• The derivative of the **outer function** \( f(u) \) at \( u = h(x) \) is noted as \( f'(h(x)) \).
• The derivative of the **inner function** \( h(x) \) with respect to \( x \) is \( h'(x) \).

Therefore, in our specific problem to find the derivative of \( f(f(x)-1) \), the original function \( f(x)-1 \) serves as the inner function and the derivative we are looking for requires us to apply these principles systematically.

Understanding Differential Calculus

Differential calculus is a branch of mathematics that studies how functions change when their inputs change.
It is primarily concerned with the concept of a derivative, which provides a way to calculate the rate at which a quantity changes. This is essential for problems involving motion, growth, and various other changes.

Key components of differential calculus include:

• **Derivatives**: Measures the rate of change of a function. For example, in physics, it is often used to find velocity and acceleration.
• **Differential equations**: Equations that relate a function with its derivatives, crucial for understanding dynamic systems.
• **Chain Rule**: A fundamental tool that allows us to differentiate compositions of functions.

In solving our exercise, differential calculus helps by providing the methods required to derive the complex relation of a composite function.
The derivative we calculated allows us to determine how the function \( f(f(x)-1) \) changes at a specific point, which in this case is at \( x=0 \). This illustrates the power of differential calculus in analyzing and solving real-world problems.