Problem 44 - Calculus Single Variable

Question

Use the specified value of \(c\) and the given information about \(f\) and \(g\) to compute \((g \circ f)^{\prime}(c)\). \(g(-5)=-6, \quad g^{\prime}(-6)=12, \quad g^{\prime}(-5)=-7, \quad f(-6)=-5\) \(f^{\prime}(-5)=1 / 3, f^{\prime}(-6)=-9, c=-6\)

Step-by-Step Solution

Answer

The derivative \\( (g \circ f)'(-6) = 63 \\\).

Step 1: Understand the Chain Rule

The problem asks for the derivative of a composite function at a point, specifically \( (g \circ f)'(c) \). We will use the Chain Rule, which states that \( (g \circ f)'(x) = g'(f(x)) \cdot f'(x) \). Our goal is to find this expression evaluated at \( c = -6 \).

Step 2: Evaluate f(c)

First, we need to find \( f(c) \). Since \( c = -6 \), we have \( f(-6) \). From the given information, \( f(-6) = -5 \).

Step 3: Calculate g'(f(c))

Next, calculate \( g'(f(c)) \). We found \( f(c) = -5 \), so we need \( g'(-5) \). According to the problem, \( g'(-5) = -7 \).

Step 4: Find f'(c)

Now, find \( f'(c) \). Since \( c = -6 \), we find \( f'(-6) \). From the information provided, \( f'(-6) = -9 \).

Step 5: Apply the Chain Rule

Use the Chain Rule formula: \( (g \circ f)'(c) = g'(f(c)) \cdot f'(c) \). Substitute the known values: \( (g \circ f)'(-6) = g'(-5) \cdot f'(-6) = -7 \cdot (-9) \).

Step 6: Compute the Result

Calculate the expression: \( -7 \cdot (-9) = 63 \). Therefore, \( (g \circ f)'(-6) = 63 \).

Key Concepts

Derivative of composite functions

When dealing with the derivative of composite functions, the chain rule is an essential tool to employ. Composite functions typically involve two functions combined into one function, written as \( (g \circ f)(x) = g(f(x)) \). To find the derivative of this type of function at a particular point, we rely on the chain rule. The rule is expressed as:

• \( (g \circ f)'(x) = g'(f(x)) \cdot f'(x) \)

This formula tells us to multiply the derivative of the outer function \( g \) at \( f(x) \) by the derivative of the inner function \( f \) at \( x \). This approach systematically breaks down a potentially complex differentiation problem into more manageable pieces. For example, in our exercise, we determined \( f(-6) = -5 \) and calculated \( g'(-5) \) and \( f'(-6) \) to find \( (g \circ f)'(-6) = 63 \). Understanding how to apply the chain rule effectively allows us to navigate through derivative problems involving nested functions efficiently.

Function composition

Function composition involves applying one function to the result of another function. Simply put, if you have two functions, \( f \) and \( g \), the composition \( (g \circ f)(x) \) means you first apply \( f \) to \( x \), and then apply \( g \) to the result of \( f(x) \). This structured process can be helpful to solve complex problems by breaking them down into simpler steps. In our exercise, to find \( f(c) \) when \( c = -6 \), we computed \( f(-6) = -5 \). This step of isolating \( f(x) \) value is crucial as it acts as the input for the next function \( g \). Without correctly identifying the output of \( f \) at \( x \), you could not compute \( g(f(x)) \). Function composition is a backbone concept in higher mathematics and is a stepping stone to understanding more advanced topics like function transformations and inversions.

Differentiation techniques

Differentiation is one of the core techniques in calculus that's used to find the rate at which a function is changing at any given point. There are several methods employed in differentiation, but when it comes to composite functions, the chain rule is unrivaled. Other techniques like the product rule and quotient rule are more suited for functions that are multiplied or divided. In dealing with our problem, understanding the derivatives and applying differentiation techniques properly involved checking known values: \( g'(-5) \) and \( f'(-6) \). These specifics highlight what aspects of each function we need to focus on. Employing differentiation appropriately allows us to understand and model dynamic systems, predict trends, and solve real-world problems which involve changing parameters. Mastering differentiation requires knowing when and how to use each rule, especially when multiple rules could be applied to reach the solution.