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Let F(x) = f(x7) and G(x) = (ƒ(x))¹ . You also know that a = 7, ƒ(a) = 2, ƒ'(a) = 15, ƒ'(a¹) Then F'(a): = and G'(a) = = = 11

Let F(x) = f(x7) and G(x) = (ƒ(x))¹ . You also know that a = 7, ƒ(a) = 2, ƒ'(a) = 15, ƒ'(a¹) Then F'(a): = and G'(a) = = = 11

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Let \( F(x) = f(x^7) \) and \( G(x) = (f(x))^7 \). You also know that \( a^6 = 7 \), \( f(a) = 2 \), \( f'(a) = 15 \), \( f'(a^7) = 11 \). Then \( F'(a) = \boxed{} \) and \( G'(a) = \boxed{} \).
Transcribed Image Text:Let \( F(x) = f(x^7) \) and \( G(x) = (f(x))^7 \). You also know that \( a^6 = 7 \), \( f(a) = 2 \), \( f'(a) = 15 \), \( f'(a^7) = 11 \). Then \( F'(a) = \boxed{} \) and \( G'(a) = \boxed{} \).
The image presents a problem involving calculus and differentiation, specifically using the chain rule. **Problem Statement:** Let \[ f(x) = (x^3 + 4x + 4)^4 \] Calculate the derivative \( f'(x) \). **Solution Provided:** \[ f'(x) = 4(x^3 + 4x + 4)^3 (3x^2 + 4) \] This expression uses the chain rule to find the derivative of the function. The outer function is \((u)^4\) where \(u = x^3 + 4x + 4\), and the inner function is \(x^3 + 4x + 4\). To further explain: - The derivative of \((u)^4\) with respect to \(u\) is \(4(u)^3\). - The derivative of the inner function \(x^3 + 4x + 4\) with respect to \(x\) is \(3x^2 + 4\). Thus, multiplying these derivatives together gives the total derivative, demonstrating the application of the chain rule. **Next Task:** Calculate \( f'(3) \), which involves substituting \( x = 3 \) into the derivative expression.
Transcribed Image Text:The image presents a problem involving calculus and differentiation, specifically using the chain rule. **Problem Statement:** Let \[ f(x) = (x^3 + 4x + 4)^4 \] Calculate the derivative \( f'(x) \). **Solution Provided:** \[ f'(x) = 4(x^3 + 4x + 4)^3 (3x^2 + 4) \] This expression uses the chain rule to find the derivative of the function. The outer function is \((u)^4\) where \(u = x^3 + 4x + 4\), and the inner function is \(x^3 + 4x + 4\). To further explain: - The derivative of \((u)^4\) with respect to \(u\) is \(4(u)^3\). - The derivative of the inner function \(x^3 + 4x + 4\) with respect to \(x\) is \(3x^2 + 4\). Thus, multiplying these derivatives together gives the total derivative, demonstrating the application of the chain rule. **Next Task:** Calculate \( f'(3) \), which involves substituting \( x = 3 \) into the derivative expression.
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