Problem 2 - Calculus Early Transcendentals

Question

Given that $$ f^{\prime}(9)=5, g(2)=9 \quad \text { and } \quad g^{\prime}(2)=-3 $$ find \((f \circ g)^{\prime}(2)\)

Step-by-Step Solution

Answer

The value of \((f \circ g)^{\prime}(2)\) is \(-15\).

Step 1: Identify the Formula

To find the derivative of a composite function \((f \circ g)(x)\), we use the chain rule. The chain rule states that \((f \circ g)^{\prime}(x) = f^{\prime}(g(x)) \cdot g^{\prime}(x)\). We will use this formula to find \((f \circ g)^{\prime}(2)\).

Step 2: Calculate \(g(2)\)

We need to determine \(g(2)\), because it is required for evaluating \(f^{\prime}(g(x))\) in the chain rule. The problem statement gives us \(g(2) = 9\).

Step 3: Calculate \(f^{\prime}(g(2))\)

Next, use the result from Step 2 to find \(f^{\prime}(g(2))\). Since \(g(2) = 9\), it follows that we need \(f^{\prime}(9)\). The problem states that \(f^{\prime}(9) = 5\).

Step 4: Use \(g^{ ext{prime}}(2)\) for Chain Rule

We also need \(g^{\prime}(2)\) to complete the chain rule calculation. The problem gives this as \(g^{\prime}(2) = -3\).

Step 5: Apply Chain Rule

Substitute \(f^{\prime}(g(2)) = f^{\prime}(9) = 5\) and \(g^{\prime}(2) = -3\) into the chain rule formula: \((f \circ g)^{\prime}(2) = 5 \cdot (-3) = -15\).

Key Concepts

Composite Functions

A composite function combines two or more functions to create a new one. When working with composite functions, you're essentially substituting the output of one function into another. Think of it as linking two processes where the result of one acts as the input for the next.
In notational terms, if you have two functions, say \( f(x) \) and \( g(x) \), their composite function is represented as \( (f \, \circ \, g)(x) \), meaning \( f(g(x)) \). This simply means you first apply \( g(x) \) and then apply \( f \) to the result of \( g(x) \). For example, if \( g(x) = x^2 \) and \( f(x) = x + 3 \), then \( (f \, \circ \, g)(x) = (x^2 + 3) \).
Understanding composite functions is crucial as they frequently appear in both basic and higher mathematical contexts, knitting together separate components into a unified whole.

Derivative Calculation

Calculating the derivative of a composite function requires understanding and utilizing the chain rule. The chain rule is a vital formula used in differentiation that helps find the derivative of composite functions.
The essence of the derivative calculation in this context involves two steps:

• First, find the derivative of the outer function \( f \) at the value given by the inner function \( g(x) \). This means evaluating \( f'(g(x)) \).
• Second, multiply the result by the derivative of the inner function, \( g'(x) \).

By combining these two derivatives, you get the derivative of the composite function itself:\[(f \, \circ \, g)'(x) = f'(g(x)) \cdot g'(x)\]
For instance, if you know \( g(2) = 9 \), \( f'(9) = 5 \), and \( g'(2) = -3 \), substituting these values into the chain rule gives you \( (f \, \circ \, g)'(2) = 5 \cdot (-3) = -15 \). This computation is a direct application of the chain rule.

Differentiation Techniques

Mastering differentiation techniques, such as using the chain rule effectively, plays a pivotal role in solving problems involving composite functions. Differentiation, in calculus, is about finding how a function changes when its input changes; essentially, it provides the slope of the function at any given point.
Some key techniques include:

• Power Rule: This helps in differentiating simple polynomials. If \( f(x) = x^n \), then its derivative is \( f'(x) = nx^{n-1} \).
• Product Rule: Used when differentiating the product of two functions. If \( u(x) \) and \( v(x) \) are functions, then \( (uv)' = u'v + uv' \).
• Quotient Rule: When differentiating a quotient of two functions. For functions \( u(x) \) and \( v(x) \), \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).

The chain rule, as described, is crucial when the function involves composition. By understanding these techniques, you can tackle various types of derivative problems more easily and accurately.