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If $$F ( x ) = f ( g ( x ) ) ,$$ where $$f ( - 2 ) = 8 , f ^ { \prime } ( - 2 ) = 4 , f ^ { \prime } ( 5 ) = 3$$ $$g ( 5 ) = - 2 ,$$ and $$g ^ { \prime } ( 5 ) = 6 ,$$ find $$F ^ { \prime } ( 5 )$$

Short Answer

Expert verified
The derivative \( F'(5) = 24 \).

Step by step solution

01

Understand the Chain Rule

To find the derivative of a composite function \( F(x) = f(g(x)) \), we use the chain rule. The chain rule states: \( F'(x) = f'(g(x)) \times g'(x) \). This means we need to evaluate \( f' \) at \( g(x) \) and multiply by \( g'(x) \).
02

Identify the Values Needed

We need \( f'(g(5)) \) and \( g'(5) \). Given \( g(5) = -2 \), we can substitute to find \( f'(g(5)) = f'(-2) \). We are also given \( f'(-2) = 4 \) and \( g'(5) = 6 \).
03

Substitute and Compute Derivative

Now substitute the values into the chain rule formula: \( F'(5) = f'(-2) \times g'(5) \). Substituting the values gives us: \( F'(5) = 4 \times 6 = 24 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Composite Functions
Composite functions involve combining two or more functions to create a new function. When we say that a function like \( F(x) = f(g(x)) \) is a composite function, it simply means that \( g(x) \) is an input to another function \( f \). This is much like a chain, where the output of one link becomes the input to another.

Understanding how composite functions work is crucial when dealing with calculus problems involving the chain rule. In this context, \( F(x) \) is the outer function, while \( g(x) \) is the inner. Whenever you encounter composite functions, think of it as a process or sequence through which you pass numbers, transforming them at each stage according to the respective functions.
  • Outer Function: The function that is applied last. In \( f(g(x)) \), \( f \) is the outer function.
  • Inner Function: The function that is applied first. In \( f(g(x)) \), \( g \) is the inner function.
The ability to distinguish between these functions is essential when applying differentiation techniques such as the chain rule.
Derivative Calculation
Calculating derivatives is a core part of calculus, and it becomes even more interesting when dealing with composite functions. The derivative of a function gives the rate at which the function value changes with respect to changes in the variable it depends upon.

When tasked with finding the derivative of a composite function \( F(x) = f(g(x)) \), you'll often use the chain rule. This involves computing the derivative of the outer function, \( f \), at the point of the inner function \( g(x) \), and then multiplying by the derivative of the inner function \( g(x) \).
  • Start by finding \( f'(g(x)) \): This involves evaluating the derivative of \( f \) at \( g(x) \).
  • Determine \( g'(x) \): This is simply the derivative of the inner function \( g \).
Combining these, the overall derivative of \( F \) with respect to \( x \) can be found using the formula \( F'(x) = f'(g(x)) \times g'(x) \). This formula is the quintessential step in the chain rule.
Differentiation Technique
The chain rule is a powerful differentiation technique used in calculus for finding the derivatives of composite functions. It essentially helps break down complex derivative calculations into simpler steps.

Begin by identifying the inner and outer functions, which is critical for applying the chain rule. Once you have these, the chain rule facilitates finding the derivative of the composite function \( F(x) = f(g(x)) \) by linking the derivatives of the constitutive functions.
  • Apply the chain rule: Using the formula \( F'(x) = f'(g(x)) \times g'(x) \), this technique allows the multiplication of the derivative of the outer function evaluated at the inner function, by the derivative of the inner function itself.
  • This method is particularly useful for complex functions where direct derivation would be cumbersome or impossible.
Through practice, the chain rule becomes an intuitive step for resolving derivative calculations in functions that are compounded or layered, paving the way for a deeper understanding in calculus.

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