Math Problem Statement

Suppose that f(1) = 1, f'(1) = 3, g(1) = 1, and g'(1) = -8. If F(x) = (f ∘ g)(x) and G(x) = (g ∘ f)(x), find F'(1) + G'(1).

Solution

The problem in the image involves applying the chain rule to find the derivatives F(1)+G(1)F'(1) + G'(1), where:

  • F(x)=(fg)(x)=f(g(x))F(x) = (f \circ g)(x) = f(g(x)),
  • G(x)=(gf)(x)=g(f(x))G(x) = (g \circ f)(x) = g(f(x)),
  • f(1)=1f(1) = 1,
  • f(1)=3f'(1) = 3,
  • g(1)=1g(1) = 1,
  • g(1)=8g'(1) = -8.

Let's break this down step-by-step to compute the derivatives.

Step 1: Compute F(x)F'(x)

Using the chain rule for F(x)=f(g(x))F(x) = f(g(x)): F(x)=f(g(x))g(x).F'(x) = f'(g(x)) \cdot g'(x). Substitute x=1x = 1: F(1)=f(g(1))g(1).F'(1) = f'(g(1)) \cdot g'(1). We know:

  • g(1)=1g(1) = 1,
  • f(1)=3f'(1) = 3,
  • g(1)=8g'(1) = -8.

Thus: F(1)=f(1)g(1)=3(8)=24.F'(1) = f'(1) \cdot g'(1) = 3 \cdot (-8) = -24.

Step 2: Compute G(x)G'(x)

Using the chain rule for G(x)=g(f(x))G(x) = g(f(x)): G(x)=g(f(x))f(x).G'(x) = g'(f(x)) \cdot f'(x). Substitute x=1x = 1: G(1)=g(f(1))f(1).G'(1) = g'(f(1)) \cdot f'(1). We know:

  • f(1)=1f(1) = 1,
  • g(1)=8g'(1) = -8,
  • f(1)=3f'(1) = 3.

Thus: G(1)=g(1)f(1)=(8)3=24.G'(1) = g'(1) \cdot f'(1) = (-8) \cdot 3 = -24.

Step 3: Compute F(1)+G(1)F'(1) + G'(1)

Now, sum the two derivatives: F(1)+G(1)=24+(24)=48.F'(1) + G'(1) = -24 + (-24) = -48.

Final Answer:

The value of F(1)+G(1)F'(1) + G'(1) is 48-48.

Let me know if you need further clarification or details on any step!


Here are some related questions you might find useful:

  1. What are the key differences between the chain rule and the product rule in calculus?
  2. How does the chain rule apply to functions with multiple variables?
  3. Can you explain the concept of higher-order derivatives and how to compute them using the chain rule?
  4. What is implicit differentiation, and how is it connected to the chain rule?
  5. How would you approach a problem where the chain rule needs to be applied multiple times?

Tip: The chain rule is crucial when differentiating compositions of functions, so practice breaking down complex expressions to master it.

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Math Problem Analysis

Mathematical Concepts

Calculus
Chain Rule
Function Composition

Formulas

F'(x) = f'(g(x)) ⋅ g'(x)
G'(x) = g'(f(x)) ⋅ f'(x)
Chain Rule: d/dx [f(g(x))] = f'(g(x)) ⋅ g'(x)

Theorems

Chain Rule

Suitable Grade Level

Undergraduate Calculus